I'm attempting to prove the following statement:
Let $X$ be a compact metrizable space. Let $F \subseteq X$ be a finite set in $X$. Then $X-F$ is compact iff every $x \in F$ is isolated.
Proof Attempt:
($\rightarrow)$ Suppose $X-F$ is compact. Assume by contradiction that there is some $z \in F$ that is not isolated. The given any open set $O$ with $z \in O$, we must have $O\cap F$ infinite as $X$ is T1 and $z$ is a limit point of $F$; a contradiction as $F$ is finite. (I'm worried about not using the hypothesis here)
($\leftarrow$) Suppose $F \subseteq X$ is finite and contains only isolated points. I aim to show that $X-F$ is closed and therefore compact as closed subsets of compact spaces are compact... (I'm not sure how to proceed from here)
You assumed that, given $x\in F$, “$x$ is an isolated points” means “$x$ is an isolated point of $F$”, whereas it actually means that $x$ is an isolated point of $X$.
$\implies$ Suppose that $X\setminus F$ is compact. Then $X\setminus F$ is closed and therefore $F$ is an open subset of $X$. But, in a metric space, a set which is both finite and open must consist of isolated points only.
$\Longleftarrow$ Let $\mathcal U$ be an open cover of $X\setminus F$. If $F=\{p_1,\ldots,p_n\}$, then $\mathcal U\cup\{p_1\}\cup\{p_2\}\cup\cdots\cup\{p_n\}$ is an open cover of $X$. Therefore, it has a finite subcover $\{U_1,U_2,\ldots,U_N\}$. But then$$\{U_1,U_2,\ldots,U_N\}\setminus\bigl\{\{p_1\},\{p_2\},\ldots,\{p_n\}\bigr\}$$ is a finite subcover of $\mathcal U$.