Countable Basis and an Uncountable Set 
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? 
 A: It cannot be the case that $\mathcal{B}_x = \mathcal{B}_y$ for $x \neq y$ in a $T_0$ space: in that case some $O$ exists such that $x \in O, y \notin O$ (or the other way round) and no element of $\mathcal{B}_y$ can be contained in $O$ but some element of $\mathcal{B}_x$ must be, so $\mathcal{B}_x \neq \mathcal{B}_y$.
Your approach should be changed. 
Let $N$ be the set of points of $A$ that are not limit points of $A$ (the uncountable set).
So for each $x \in N$ we have some open set $O$ such that $x \in O$ and $O \cap A = \{x\}$, and we can pick a base element $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq O$ and so $B_x \cap A = \{x\}$. For $x \neq y$ this implies $B_x \neq B_y$. So the map $x \to B_x$ is 1-1 from $N$ to $\mathcal{B}$ and the latter is a countable set, so $N$ is at most countable. It follows that $A \setminus N$ is the uncountable subset of $A$ that are limit points.
A: Following Henno's answer I have an explicit counter-example.
Take $\mathbb{R}$ with the standard topology, and $\mathcal{B}$ be all the intervals $(q,p)$ for $q,p\in\mathbb{Q}$, this is a countable basis.
For every $x$, $\mathcal{B}_x$ is the set of intervals $(q,p)$ for $q<x<p$. Which is countable.
Take $A=(0,1)$, then for every $x\in A$ you have that $\mathcal{B}_x\subseteq \mathcal{B}$ (obviously). And $\bigcup_{x\in A} \mathcal{B}_x \subseteq \mathcal{B}$, but not $\mathcal{B}_x = \mathcal{B}_y$ because if $x\not = y$ then you can take $q,p$ sufficiently close to $x$ such that $(q,p)$ contains $x$ but doesn't contain $y$, in particular $(q,p)\in \mathcal{B}_y$ but not in $\mathcal{B}_x$.
