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 $\epsilon>0$, $B(x;\epsilon)\cap C$ 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.