$C$ is an uncountable set. Show that $B(x,\epsilon)\cap C$ is uncountable. Suppose $X$ is a separable metric space, and $C$ is an uncountable subset of $X$. Prove that there is a point $x \in C$ such that for each $\epsilon&gt;0$, $B(x;\epsilon)\cap C$ is uncountable.


I missed first couple of classes so I don't really know what is the general strategy to prove something is uncountable (or countable)? The only thing that comes to my mind is that, since $X$ is separable, then for any open cover, it has a countable sub cover (may not be finite though), and I was hoping I could get a contradiction out of here? 

I tried contradiction, but that would means for every $x\in C$, there is a corresponding $\epsilon(x)$ such that $B(x;\epsilon(x))\cap C$ is countable. But I cannot deduce that $C$ is then countable because that intersection may be empty.

Any idea?
Even more is true, and not much harder to prove.

HINT: Let $\mathscr{B}$ be a countable base for $X$. Let $$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap C\text{ is countable}\}\;.$$ 

Let $C_0=C\setminus\bigcup\mathscr{B}_0$. Show that $C_0$ is actually an uncountable set such that $B\cap C_0$ is uncountable for every open nbhd $B$ of every $x\in C_0$.
Initially I was proving existence of $x\in X$ since I misread the problem this way. Now I made the small changes to prove existence of such $x\in C$.

Let $\{x_n\}\subset C$ be dense in $C$. Assume no such $x\in C$ exists. Then there are $r_n&gt;0$, largest, such that $B(x_n,r_n)\cap C$ is countable because, in particular, none of the $x_n$ is a candidate for $x$. 

But $\bigcup_n B(x_n,r_n)\supset C$. This implies that $C$ is countable.



Assume $y\in C\setminus\bigcup_n B(x_n,r_n)$ (in particular $y\neq x_n$ for all $n$). Let $r&gt;0$ be such that $B(y,r)\cap C$ is countable, which by the assumption exists. Let $x_n$ be inside $B(y,r/4)$. Then $r_n\leq d(y,x_n)&lt;r/4$. But $B(x_n,3r/4)\subset B(y,r)$ and therefore $B(x_n,3r/4)\cap C$ is countable contradicting the choice of $r_n$ (since $3r/4&gt;r/4&gt;r_n$).

Therefore $\bigcup_n B(x_n,r_n)\supset C$ as claimed.



Existence of the largest $r$:

Under the assumptions that $C$ is uncountable, and that $x$ is such that there is $\epsilon&gt;0$ such that $B(x,\epsilon)\cap C$ is countable. Let $r:=\sup\{s\in\mathbb{R}^+:\ B(x,s)\cap C\text{ is countable}\}$. Since $B(x,r)=\bigcup_{n\in\mathbb{N}}B(x,r-1/n)$ we have that $B(x,r)\cap C=\bigcup_{n\in\mathbb{N}}\left(B(x,r-1/n)\cap C\right)$ is countable.