I'm trying to prove this theorem:
Let $(X,d)$ be a complete metric space, with no isolated points, prove X is uncountable.
I've heard that this can be proved using Baire's category theorem but we haven't learned that theorem in my course so I'm looking for a proof that does not use it.
I started by assuming this is not true and marking $X = \{x_n|n\in \Bbb{N}\}$. Knowing that X is complete I can use the fact that any Cauchy series converges to some element in X.
I think that an argument as the following might work:
Suppose that $X$ is countable, then you can index its elements by $\mathbb{N}$, as $\{x_n\}_{\mathbb{N}}$.
Consider a ball $B(x_1,r)$, then you have infinitely many $x_i$s in the ball (because in $X$ there are not isolated points). Pick the first $i_1>1$ such that $x_{i_1}\in B(x_1,r)$, and a ball $B(x_{i_1}, r_1)$ contained in $B(x_1,r)$ such that $x_1$ is not in $B(x_{i_1},r_{i_1})$. Inductively you can construct in this way a sequence of $x_{i_n}$s such that $x_{i_k}\in B(x_{i_n},r_n)$ for $k\geq n$, and such that $x_{i_{n+1}}$ is the element with the first index $j$ such that $x_j\in B(x_{i_n},r_n)$. Moreover you can impose that $x_{i_n}$ is not in $B(x_{i_{n+1}},r_{n+1})\subset B(x_{i_n},r_n)$ and choose the $r_n$s in a way that they tend to $0$.
The sequence then is clearly Cauchy (it is contained in balls that get smaller and smaller), but it has no limit. In fact if $x_l$ is the limit of the sequence, then $l$ can't be one of the $i_n$s by the construction of the sequence. Then you have that $i_{j}<l<i_{j+1}$ for some $j$, but since you must have $x_l\in B(x_{i_j},r_j)$ (it is the limit) this is not possible by the construction of the sequence (remember that $i_{j+1}$ is the first index such that the corresponding element is contained in the ball). This proves the claim by contradiction.
According to the wikipedia page on Perfect Sets, the Cantor set can be continuously embedded in any complete metric space w/o isolated points. This is an extension of Cantor's theorem that any non-empty perfect set of real numbers has cardinality of the continuum. Cantor's theorem is from 1883, predates BCT and has an indepenedent proof. I think the extension of building a Cantor set inside any complete metric space w/o isolated points can be done without BCT, but facially it seems that this approach is a lot more complicated. In other words, using the BCT is the short way to prove this.
Of course it should be assumed that the space is nonempty. Given that, I'll first produce a countable subset of $X$, indexed by the finite sequences of 0s and 1s. Start by letting $x_\varnothing$ be an arbitrary point. Once $x_s$ has been defined for all of the $2^n$ sequences $s$ of length $n$, pick for each of these points $x_s$, an open ball $B_s$ around it, small enough so that these $2^n$ balls have pairwise disjoint closures, have diameters $<2^{-n}$, and are $\subseteq B_t$ for every proper initial segment $t$ of $s$. Then, since no $x_s$ is isolated, let $x_{s0}$ and $x_{s1}$ be two points in $B_s$ distinct from each other and from $x_s$. This completes the inductive construction of the points $x_s$ for all finite sequences $s$ of 0s and 1s.
Now given any infinite sequence $Q$ of 0s and 1s, consider the points $x_s$ where $s$ is an initial segment of $Q$. Check that these points constitute a Cauchy sequence (because the diameters of the balls $B_s$ are small), and let $y_Q$ be its limit (because the space $X$ is complete).
Finally, because of the nesting and disjointness of the balls chosen at the various stages, the points $y_Q$ are distinct for distinct sequences $Q$. Since there are $2^{\aleph_0}$ choices for $Q$, there are $2^{\aleph_0}$ points $y_Q$.
