The set $E$ cannot be open. If open, then for some $(a,b)\subset E \implies c\in(a,b) $ is a limit point of $E$, a contradiction. Hence, the set must be closed. If $E=\emptyset$, then the conclusion is evident.
The set $E$ if bounded, cannot be infinite. Since, by Bolzano-Weierstrass theorem, $E$ must have a limit point, and being closed, that point must be in $E$.
The conclusion in this case is also established.
Now, let $E$ be unbounded. Assume it is uncountable. We can pick a bounded uncountable subset $C$ of $E$. Hence by Bolzano-Weierstrass theorem, it must contain the limit point of some subsequence of an infinite bounded sequence in $C$. $E$ being closed, the limit point must belong to $E$. Therefore, we get a contradiction.
The theorem is established.
Is the proof correct? I looked up this question here but I did not get a satisfactory (they were a bit harder to grasp) answer. Please do check.
Best Answer
$E$ is a subset of a separable metric space. So, as a metric space in its own right, $(E,d)$ is separable. By hypothesis, $(E,d)$ has no accumulation points, therefore $E$ is the only dense subset of $(E,d)$: otherwise $x\in E\setminus(\text{a dense subset})$ would be an accumulation point of the dense subset. Since $(E,d)$ is separable, $E$ must be countable.