Every closed subset of $\Bbb R$ has countably many isolated points.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $A \subseteq \Bbb R$ be a closed set and $I$ be the set of all isolated points of $A$.
If $a \in I$, then there exists $\delta > 0$ such that $x \neq a$ and $|x-a|<\delta$ implies $x \notin A$. Since $\Bbb Q$ is dense in $\Bbb R$, there exist $r,s \in \Bbb Q$ such that $a-\delta<r<a<s<a+\delta$.
As a result, if $a \in I$, then there exist $r,s \in \Bbb Q$ such that $a$ is the only element of $A$ in $(r,s)$.
Let $\langle r_n,s_n \rangle_{n \in \Bbb N}$ be an enumeration of the set $\{\langle r,s \rangle \mid r,s \in \Bbb Q\}$.
We define a mapping $\mathcal F:I \to \Bbb N$ by $$\mathcal F(a) = \min \{n \in \Bbb N \mid a \text{ is the only element of } A \text{ in } (r_n,s_n)\}$$
Clearly, $\mathcal F$ is injective and thus $|I| \le |\Bbb N| = \aleph_0$. Hence $I$ is countable.
Best Answer
Your proof looks correct for $\mathbb{R}$ with its standard topology. Although, as Lee Mosher pointed out, it could be simplified.
To see that your statement is something that depends on our knowledge of $\mathbb{R}$ with its usual metric, and it is not necessarily true in general topological spaces, or even metric spaces, consider the same set, the good old $\mathbb{R}$ with the discrete topology this time. That is, consider $\mathbb{R}$ equipped with the discrete metric:
$$d(x,y) = \begin{cases}0 & x=y \\ 1 & x\neq y\end{cases}$$
Then all points of $\mathbb{R}$ will become isolated points (take $0<\delta<1$), proving that your statement is not necessarily true about $\mathbb{R}$ with a topology other than the one coming from the Euclidean distance and it's a very specific statement in real analysis.
Also, your statement is true about separable metric spaces, i.e. any metric space that has a countable, dense subset. And the proof is similar to what you wrote taken Lee Mosher's comment into account.