Let $X$ be a compact metric space with no isolated points (so every point is a limit point). Show that $X$ is uncountable.
I think I can show this by showing that $X$ has cardinality at least that of the Cantor Set. Essentially, begin with two disjoint nonempty open sets $U_0$ and $U_1$ of $X$. Then choose disjoint nonempty open sets $U_{00}, U_{01}$ of $U_0$ and $U_{10}, U_{11}$ of $U_1.$ Though I'm not really sure how to show that we can always choose such sets. The axiom of choice is likely involved here.
However, if we can always choose such sets, then after choosing the $2^n$ sets $U_{0\cdots 0},\cdots, U_{1\cdots 1},$ for each of the $2^n$ sets $U_{j_1\cdots j_n}$, we can choose two sets $U_{j_1\cdots j_n0}, U_{j_1\cdots j_n1}$ that are disjoint and nonempty in $U_{j_1\cdots j_n}.$ How would one formally show that this results in a "copy" of the Cantor set in $X$?
Edit: I do not want to use the Baire Category theorem to solve this problem.
Best Answer
Hint.
Your answer is basically good. But (a) you want closed sets (not open) so that, if $j_1j_2\dots$ is an infinite sequence, you can conclude that $$ \bigcap_{n=1}^\infty U_{j_1j_2\dots j_n} \tag1$$ is nonempty. And (b) you want to arrange that the diameters if $U_{j_1j_2\dots j_n}$ are ${} < 1/n$ [or something going to $0$], so that $(1)$ is a single point.