An outline of what I understand to be a proof that $0.9999\dotso=1$:
$1/3 = 0.333\dotso$
$1/3 + 1/3 + 1/3 = 1$
$0.333\dotso + 0.333\dotso + 0.333\dotso = 0.999\dotso$
Therefore, $1 = 0.999\dotso$
This confuses me because the setup and math is strikingly similar to a proof by contradiction. Shouldn't the last line be replaced with
Therefore, since we know $1$ does not equal $0.999\dotso$, $0.333\dotso$ is not equal to $1/3$.
Why is this still considered a valid proof that $0.999\dotso = 1$?
It appears to me that there is a nexus of three misconceptions or misunderstandings here which make this question about more than just the value of $0.\overline{9}$ (or $0.999\dotso$, if you prefer):
there is confusion about what, precisely, the notation $0.999\dotso$ means,
there is confusion about the nature of the argument presented, and
there is confusion about the nature of proof by contradiction.
The goal of this answer is to address these three problems.
The problem here seems to be with the use of the ellipses ($\dotso$) throughout the argument. Ellipses are not a rigorous bit of notation, and are meant to hide details that an experienced mathematician ought to be able to fill in on their own. In this particular case, the "missing piece" is a solid understanding of what the decimal representation of a number actually means, so let's start by filling that in:
Definition: Suppose that $n$ is a nonnegative integer and that $(d_j)_{j=-n}^{\infty}$ is a sequence of integers with $$ d_j \in \{0,1,2,3,4,5,6,7,8,9\}. $$ Then define the notation $$ d_{-n} d_{-(n-1)} \dotso d_{-1} d_0 . d_1 d_2 d_3 \dotso = \sum_{j=-n}^{\infty} \frac{d_j}{10^n}. $$ That is, the concatenation of the terms of $(d_j)$ (with a period placed between $d_0$ and $d_1$) is defined to be the series on the right. Note that for any sequence $(d_j)$ of the kind described, the series on the right will converge to some nonnegative real number $x$. The notation on the right is called a decimal expansion of $x$.
The fact that the series on the right converges is a consequence of basic definitions and results in real analysis—this is not the place to get into those details; we will have to take them as given. The advantage of having a definition like the one above is that it gives meaning to all of those ellipses, and gives us a more concrete mathematical object to work with.
I will note that there are a couple of lacunae in this definition. I have very intentionally defined the notation $d_{-n}\dotso d_0.d_1\dotso$ to be a decimal expansion of some real number $x$, and not the decimal expansion. The definition makes no claim about uniqueness, and allows for the possibility that there may be multiple series of the kind described which all converge to the same real number $x$. Note, also, that the definition makes no claim about the existence of a decimal expansion for an arbitrary real number—it says, given a particular kind of sequence, we can construct a real number in terms of a decimal representation. I have not asserted that every real number has a decimal expansion (while this is true, it is not necessary for seeing what is going on here).
In light of this definition, we have the following: $$ 0.\overline{3} = \sum_{j=1}^{\infty} \frac{3}{10^j} \qquad \text{and} \qquad 0.\overline{9} = \sum_{j=1}^{\infty} \frac{9}{10^j}. $$ These series are both related to geometric series, and the limit of a geometric series is a well understood problem in mathematics. In this case, we have $$ \sum_{j=1}^{\infty} \frac{d}{10^j} = d \sum_{j=1}^{\infty} \frac{1}{10^j} = d \frac{1}{1-10} = \frac{d}{9}. $$ Therefore $$ \sum_{j=1}^{\infty} \frac{3}{10^j} = \frac{3}{9} = \frac{1}{3} \qquad \text{and} \qquad \sum_{j=1}^{\infty} \frac{9}{10^j} = \frac{9}{9} = 1. $$
The conclusion is, then, that if we properly define the notation we are using, then we obtain both of the claimed results directly from our definitions.
Alternatively, we can try to address what this "proof" is actually trying to say, and confront it on its own terms. First and foremost, I would argue that the reasoning is not very rigorous, and is meant to be a heuristic argument rather than a formal proof. It is meant to be convincing to an audience that doesn't really have all of the tools required for a more rigorous approach.
This argument is generally presented to high school students, the tools of analysis (e.g. series) are, perhaps, out of reach. Thus I might present the given argument as follows:
Using the division algorithm, we have \begin{align} 1 \div 3 &= 0 \text{ r } 1 \\ 1.0 \div 3 &= 0.3 \text{ r } 0.1 \\ 1.00 \div 3 &= 0.33 \text{ r } 0.01 \\ 1.000 \div 3 &= 0.333 \text{ r } 0.001, \end{align} where (for example) $0.3 \text{ r } 0.1$ denotes a quotient ($0.3$) and a remainder term ($0.1$). This pattern continues forever, thus we conclude that $$ \frac{1}{3} = 1 \div 3 = 0.\overline{3}. $$ It then follows that $$ 1 = 3 \times \frac{1}{3} = 3 \times (0.\overline{3}) = 0.\overline{9}. $$
There are two important gaps here:
The phrase "this pattern continues forever" is not justified. The presentation here relies on some intuition and on some magical handwaving. This is appropriate for the intended audience, as the tools required to make this rigorous come from analysis, and high school students are not expected to understand analysis.
I have not justified the equality $3 \times 0.\overline{3} = 0.\overline{9}$. Again, making this rigorous requires analysis, so skipping it here isn't fatal.
Again, keep in mind that this isn't really a proof. It is a convincing argument. We could probably turn it into a proof with some work, but once we get our hands on the appropriate tools from analysis, it turns out that there are more straight-forward proofs (see above).
In a proof by contradiction, you first assume a faulty premise, then use valid inference to deduce that a known false statement follows. For example, one of the first contradiction proofs that most students learn is one which demonstrates the infinitude of the primes.
The proof starts with a false assumption: suppose that there are only finitely many primes, say $\{ p_1, p_2, \dotsc, p_n\}$. From this false assumption, a contradiction is produced by discovering the existence of a prime which is not in the list.
In the case of the argument here, no contradiction is ever produced. You start by assuming that $$ \frac{1}{3} = 0.\overline{3}, $$ and then conclude that $$ 1 = 0.\overline{9}. $$ We have no reason to believe that $1 \ne 0.\overline{9}$, and so there is no clearly false statement at the end of the end of the argument. Unless we can prove by some other means that $0.\overline{9}$ is not $1$, then the best we can say is that if $1/3 = 0.\overline{3}$, then it must follow that $0.\overline{9} = 1$.
In a proof by contradiction, we demonstrate that a proposition is true by:
For example, in the proof by contradiction that $\sqrt{2}$ is irrational, we begin by assuming that $\sqrt{2}$ can be written in the form $p/q$, where $p$ and $q$ are in lowest terms. Then, we show that this assumption implies that $p$ and $q$ are both even, which conflicts with the assumption that $p$ and $q$ are in lowest terms.
Suppose we tried to turn your proof into a proof by contradiction. This won't work as all of the statements in your proof are actually true, but we'll put this to the side for a moment:
The point I am trying to make is that even if your proof by contradiction was valid, all it would show would be that $1/3\neq0.333\dots$
Your proof by contradiction isn't valid, though, as it is the case that $0.999\ldots=1$, and so no contradiction has been reached. Let me try to explain in more detail why $0.999\ldots=1$. We must first show that $$ \frac{1}{3}=0.333\ldots \tag{*}\label{*} $$ In order to get anywhere near showing that $\eqref{*}$ is true, we must first be clear on what $0.333\ldots$ actually means. By definition, $0.333\ldots$ is the limit of the series $$ S_n = \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \dots + \frac{3}{10^n} $$ as $n$ tends to infinity. Note that $$ \frac{1}{10}S_n=\frac{3}{100} + \frac{3}{1000} + \frac{3}{10000}+\dots+\frac{3}{10^{n+1}} \, . $$ If we subtract $\frac{1}{10}S_n$ from $S_n$, most of the terms cancel, and we are left with $$ S_n - \frac{1}{10}S_n = \frac{3}{10} - \frac{3}{10^{n+1}} \, . $$ From here, we just need to simplify and we arrive at the result: \begin{align} \frac{9}{10}S_n &= \frac{3}{10}\left(1-\frac{1}{10^n}\right) \\[5pt] S_n &= \frac{1}{3}\left(1-\frac{1}{10^n}\right) \, . \end{align} As $n$ gets larger and larger, $1/10^n$ gets closer and closer to $0$, meaning that the expression within the bracket approaches $1$. Hence, $$ \lim_{n \to \infty}S_n=\frac{1}{3} \, . $$ This establishes that $0.333\ldots$, which is defined as $\lim_{n \to \infty}S_n$, is equal to $\frac{1}{3}$. If we multiply both sides of the equation by $3$, we get that $$ 3 \cdot \lim_{n \to \infty}S_n = 1 $$ Moreover, the limit law $c \cdot \lim_{n \to \infty}S_n= \lim_{n \to \infty}c \cdot S_n$ for any constant $c$ tells us that $$ \lim_{n \to \infty}3S_n = 1 \, . $$ Note that $$ 3S_n = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots + \frac{9}{10^n} $$ and so by definition, $0.999\ldots=\lim_{n \to \infty}3S_n$. This establishes that $0.999\ldots=1$.
