Every point of the set of all uncountable limit points is a limit point A friend told me me about this problem and although maybe it is easy, I don't know how to proceed. I tried to proceed by contradiction but didn't find anything useful. I appreiate your help.

Let $A$ be a subset of $\mathbb R^k$ and let $X$ be the set of all points $x$ such that every neighborhood around $x$ contains uncountably many points of $A.$ Prove that $X$ contains no isolated points.

I should add that for me a neighborhood around $x$ is the set of all points $y$ such that $\|x-y\|<\epsilon$ for some positive real number $\epsilon.$
 A: Recall that $\Bbb R^k$ has a countable base $\mathscr{B}=\{B_n:n\in\Bbb N\}$. (For instance, we can take $\mathscr{B}$ to be the set of all products of the form $\prod_{i=1}^k(p_i,q_i)$, where $p_i,q_i\in\Bbb Q$ and $p_i<q_i$ for $i=1,\ldots,k$.) 
Now suppose that $x$ is an isolated point of $X$; then $x$ has an open nbhd $U$ such that $U\cap X=\{x\}$. Let $y\in U\setminus\{x\}$; then $y\notin X$, so $y$ has an open nbhd $U_y$ such that $A\cap U_y$ is countable. $\mathscr{B}$ is a base for the topology on $\Bbb R^k$, so there is an $n(y)\in\Bbb N$ such that $y\in B_{n(y)}\subseteq U_y$. In short, for each $y\in U\setminus\{x\}$ there is an $n(y)\in\Bbb N$ such that $y\in B_{n(y)}$, and $A\cap B_{n(y)}$ is countable. 
Let $M=\big\{n(y):y\in U\setminus\{x\}\big\}$, and note that $B_m\cap A$ is countable for each $m\in M$. Let $V=\bigcup_{m\in M}B_m$; $M$ is clearly countable, and 
$$A\cap V=A\cap\bigcup_{m\in M}B_m=\bigcup_{m\in M}(A\cap B_m)$$
is therefore the union of countably many countable sets and hence itself countable. But $A\cap U$ is uncountable (since $U$ is an open nbhd of $x$), so $A\cap(U\setminus\{x\})$ is uncountable, and the construction of $V$ ensures that $U\setminus\{x\}\subseteq V$, so $A\cap V$ is uncountable. This contradiction shows that $x$ cannot be isolated in $X$ after all.
The basic idea is very simple, though it may have got lost in the technical details. If $x$ is isolated in $X$, it has an open nbhd $U$ that contains no other point of $X$. This means that each point of $U$ except $x$ itself has an open nbhd (and hence a basic open nbhd) containing only countably many points of $A$. There are only countably many basic open sets altogether, so the union of these basic open nbhds of points of $U\setminus\{X\}$ still contains only countably many points of $A$. But that union is all of $U$ except possibly the point $x$ itself, so $U$ contains only countably many points of $A$, contradicting the choice of $x$ as an element of $X$.
A: Let $x\in\mathbb R^k$ be such that $\exists r>0$ such that the open ball $B_r(x)$ intersects $X$ in at most its center $x.$ Then for each $y\in B_r(x)-\{x\}$ there is some $\varepsilon(y)>0$ such that $B_{\varepsilon(y)}(y)\cap A$ is at most countable. Then $\{B_{r/2^2}(x)\}\bigcup\{B_{\varepsilon(y)}(y)\}_{0<\|y-x\|\leqslant r/2}$ is an open cover of the closed ball $\overline{B_{r/2}(x)}.$ Since every closed ball is compact in $\mathbb R^k,$ then $(B_{r/2}(x)-B_{r/2^2}(x))\cap A$ is at most countable. Applying the same process again, we conclude that $(B_{r/2^2}(x)-B_{r/2^3}(x))\cap A$ is at most countable and hence $(B_{r/2}(x)-B_{r/2^3}(x))\cap A$ is at most countable. Thus if we keep applying this process we conclude that $(B_{r/2}(x)-B_{r/2^n}(x))\cap A$ is at most countable for all integers $n\geqslant2.$ Therefore $(B_{r/2}(x)-\{x\})\cap A$ is at most countable, which implies that $x\notin X.$ Hence $X$ contains no isolated points.
Note. A closed set with no isolated points is called perfect.
Edit: I made a mistake in my original answer by assuming that the collection $\{B_{\varepsilon(y)}(y)\}_{0<\|y-x\|\leqslant r/2}$ is an open cover of the closed ball $\overline{B_{r/2}(x)}$ so I hope my edited answer has no flaws.
