No Limit Points $\implies$ Countable 
If $\Omega \in \mathbb{C}$ is open, show that any of its subsets without limit points is at most countable.

I have seen answers here on how to prove that no limit points implies at most countable: Need a hint: show that a subset $E \subset \mathbb{R}$ with no limit points is at most countable.
But, I was wondering if the following argument holds: Assume a set A is uncountable and has no limit points. Then, we can find an $r>0$ and $x \in A$ such that $N_{r}(x)$ contains infinitely many elements of $A$ because $A$ is uncountable [EDIT: I now realize this only works if $\Omega = \mathbb{C}$]. But then, the closure of this region is compact, so the set does have a limit point.
 A: Your  argument :
" Assume a set $A$ is uncountable and has no limit points. Then, we can find an $r>0$ and $x∈A$ such that $N_r(x)$ contains infinitely many elements of $ A $ because $A$ is uncountable. But then, the closure of this region is compact, so the set does have a limit point."
Does not work, because the set $A$ in uncountable and there is no need for some point $x$ in $A$ to have infinitely many elements in $N_r(x)$ to make it uncountable.
A: The problem here is that you don't necessarily have $N_r(x) \subseteq A$, so even if $N_r(x)$ has a limit point, that point doesn't need to be a limit point of $A$.
However, your argument can be fixed with some work:
Given $r > 0$, let $A_r \subseteq A$ be the set of all points such that the neighborhood of radius $r$ around this points contain a point of $A$ other than that point, i.e.
$$A_r = \{x \in A : |N_r(x) \cap A| \geq 2\}.$$
(should work just as well with infinitely many points)
Then because $A$ is uncountable, we get $A_r \neq \emptyset$ for all $r > 0$.
Furthermore, we have that 
$$r < s \Rightarrow A_r \subseteq A_s.$$
Using these two properties, you can argue that
$$\lim_{r \to 0} A_r$$
contains at least one limit point of $A$.
You might need to use Zorn's Lemma though...
