Let $(X,d)$ be a metric space. Let $A \subset X$. We say $x \in X$ is a condensation point of $A$ if every neighborhood of $x$ contains an uncountable number of elements of $A$. Prove every uncountable subset of $\mathbb{R}$ has a condensation point.
My current thought is proving by contradiction. Suppose no point of $A$ is a condensation point, every uncountable subset of $\mathbb{R}$ has countable number of elements around a point $x$. However, I don't know how to proceed from here.
Yes, that's a good starting point. For each $n\in\mathbb Z$, let $A_n=A\cap[n,n+1]$. Then $A=\bigcup_{n\in\mathbb Z}A_n$ and therefore some $A_n$ is uncountable too. Suppoes that $A_n$ has no accumulation point. For each $x\in[n,n+1]$, let $U_x$ be an open set such that $x\in U_x$ and that $A_x$ contains only countably many points of $A$. Since $[n,n+1]$ is compact, there is a finite set $\{x_1,x_2,\ldots x_N\}\subset[n,n+1]$ such than$$A_n\subset[n,n+1]\subset\bigcup_{j=1}^NU_{x_j}.$$This is impossible, because each $U_{x_j}$ only has countably many elements of $A$, whereas $A_n$ is uncountable.
Start with some finite interval $I$ such that $A\cap I$ is uncountable (this has to be true since $A$ is the countable union $\bigcup_{n\in\mathbb{N}}(A\cap [-n,n])$. Now proceed similar to the proof of Bolzano-Weierstrass, $I_0=I$, halving the length of interval $I$ each time: $I_{n+1}$ is either the left half $[\min I_n,(\min I_n+\max I_n)/2]$ or the right half $[(\min I_n+\max I_n)/2,\max I_n]$ so that $A\cap I_{n+1}$ is uncountable (for definiteness, say choose the left half if both are allowed). Then $\bigcap_n I_n$ is nonempty and is a singleton $\{x\}$, and obviously $x$ is a condensation point of $A$ since every neighbourhood contains $I_n$ for some $n$.
A separable metric space like $\mathbb{R}$ is Lindelöf (see the equivalence of 3. and 7. in this answer e.g.), and so if $A$ is an uncountable set and we assume that it has no condensation point, then for each $x \in \mathbb{R}$ we have an open set $U_x \ni x$ with $U_x \cap A$ countable at most (so finite or empty are possible).
We cover $\mathbb{R}$ by a countable subcover $U_x, x \in M$ (so $M$ is some countable set of points whose neighbourhoods still cover the reals) and then
$$A = \bigcup_{x \in M} (U_x \cap A)$$ would be a countable union of countable sets, which is false.
So $A$ does have a condensation point.
