Must an uncountable subset of R have uncountably many accumulation points? [duplicate]

This question is taken from problem 4.1.8 of "Real Analysis and Foundations" by Krantz

The question reads:

"Let S be an uncountable subset of $\mathbb{R}$. Prove that S must have infinitely many accumulation points. Must it have uncountably many?"

The first part of the question took some work but ended up coming out pretty smoothly; however, I'm at a complete loss as to how to go about addressing the second problem. Intuitively, I think the answer is yes.

My first attempt was to try to show that a countable number of accumulation points allowed one to order the elements of S in such a way that they are countable (ie prove the contrapositive), but I did not manage to get much further than that.

My second attempt was to show that $S-{s_{1},s_{2}...}$ where $s_{1},s_{2},...$ are countably many accumulation points of S is uncountable and thus must have an accumulation point, so S has an accumulation point that is not one of the countably infinite set. Hence, the number of accumulation points is uncountable.

My question is, is this logic valid? I would have to prove that an uncountable set minus a countable set is uncountable, which shouldn't be too difficult.

Any hints/points in the right direction/outright answers are greatly appreciated.

Solution 1:

Let $T$ be the set of elements of $S$ that are not accumulation points of $S$. Then for each $x \in T$ there exists $\epsilon_x > 0$ such that the interval $I(x,\epsilon_x) = (x-\epsilon_x,x+\epsilon_x)$ contains no other points of $S$. The intervals $\{I(x,\epsilon_x/2) | x\in T\}$ are disjoint, and each contains a rational number; so there can only be a countable number of them.

Hence $T$ is countable, and the set of accumulation points of $S$, which contains $S-T$, is uncountable.