Let $X$ have a countable basis; let $A$ be an uncountable subset of $X$. Show that uncountably many points of $A$ are limit points of $A$.
Although I find this question asked on MSE, I believe my proof is sufficiently different; and I believe it is correct, but I suppose you'll be the judge. I am, however, having trouble justifying one step, which I will point out at the end of my proof:
Suppose that $X$ has a countable basis $\mathcal{B}$ (i.e., that it is second countable, and let $A$ be some uncountable subset of $X$. Then $X$ must also be first countable which means that every point $x$ has a countable basis $\mathcal{B}_x$. Let $x \in A-E$ and $\mathcal{B}_x$ its associated basis. Note that $\bigcup_{x \in A} \mathcal{B}_x \subseteq \mathcal{B}$ is a uncountable union of countable sets contained in a countable set which is a contradiction, unless there exists a $z \in A$ such that $\mathcal{B}_y = \mathcal{B}_z$ for uncountably many $y \in A$. Letting $E$ denote the collection of such $y$, it's clear that $E$ consists of some of $A$'s limit points: for if $y \in E$, and $B \in \mathcal{B}_y$ is arbitrary, then $z \in B$ and therefore $B \cap (A - \{y\})$ is not empty.
In the MSE chatroom, Ted Shifrin informed me that means that an uncountable union of countable sets being contained in a countable set implies at least one of the sets in the union occurs uncountably many times, which I took to mean there exists a $\mathcal{B}_x$ such that $\mathcal{B}_y = \mathcal{B}_x$ for uncountably many $y \in A$. How does one prove this?