Think about what continuity intuitively is about: it means "close" input values get mapped to "close" output values. But with the constant $0$ function of course all values in the image will be close to each other no matter how far the input values are apart - it's basically as continuous as it gets.
Explanations for why your derivation is incorrect are already given in the comments multiple times so I won't go into that. There's also another answer with a solution (the so-called Dirichlet function - the indicator function of the set of rational numbers) but it doesn't really go into why this is a "natural" solution.
We've already noticed that the $0$ function is as continuous as it gets - it's perfectly smooth. So lets start with that and modify it such that it has a discontinuity. At the beginning I've said that continuity is about closeness, that we can make our output values be as close to one another as we want by simply considering small and smaller input intervals - which functions don't statisfy this property? Surely if we had a function $f$ such that $|f(x) - f(y)|$ was constant for $x \neq y$ we'd have this. So lets define exactly this kind of function. Let $x_0 \in \Bbb R$ and define
$$
f_0(x) = \begin{cases} 1, &\text{if }x = x_0 \\0, & \text{otherwise.} \end{cases}
$$
Then no matter how small a ball (so a set of the form $(x_0 - \delta, x_0 + \delta)$ for $\delta > 0$) we draw around $x_0$ there'll always be some point $x$ in it that is not equal to $x_0$ and thus we'll have $|f(x) - f(x_0)| = 1$. So the prototypical discontinuous functions are precisely the jump functions - another commonly used example are step functions like $f(x) = \begin{cases} 1, &\text{if }x > x_0 \\0, & \text{otherwise.} \end{cases}$
To add more discontinuities we can just add up functions that look like $f_0$, which of course has the same effect as adding more branches that check for equality or we replace the equality check by a single set membership check. So for some real numbers $x_0, ..., x_n$ the function $$f_1(x) = \begin{cases} 1, &\text{if }x \in \{x_0, ..., x_n \}\\0, & \text{otherwise} \end{cases}$$
will have discontinuities precisely at the points $x_0,...,x_n$. And now we just need to blow up this set of discontinuities in some way. We'll denote the function that is $1$ for all $x$ in some set $A$ and $0$ elsewhere by $\chi_A$ and call it the indicator function of $A$. Lets note some properties that we want our set $A$ to have such that $\chi_A$ will be everywhere discontinuous:
- $A$ should be infinite, because for any finite set we'll easily find some interval on which the function $\chi_A$ is identically $0$
- $A$ can't contain any interval, because otherwise $\chi_A$ would be continuous precisely on this interval
- in the same way $\Bbb R \setminus A$ can't contain any intervals, lest we'd have continuity on that.
So we're looking for an infinite set that doesn't contain any intervals itself and breaks up all possible intervals. For any such set $A$ the function $\chi_A$ will be everywhere discontinuous.In topological terms we're looking for a dense subset of $\Bbb R$.
What infinite subsets of the real numbers are there that we might consider? The simplest one that probably comes to mind are the natural numbers and the integers but clearly they both don't "break up all intervals".
The next "bigger" set that we often times deal with are the rationals $\Bbb Q$ and we can quickly check for ourselves that they do indeed have all the properties we want. Thus $\chi_{\Bbb Q}$ is everywhere discontinuous.
