Proof that a limit point compact metric space is compact. If $(X,d)$ is limit point compact, show it is compact.
I have found multiple proofs of this that first show that limit point compact implies sequential compact, which in turn implies compactness. I was wondering if there is a direct proof showing that limit point compactness implies compactness in a metric space.

Suppose $E \subseteq X$ is the limit point compact infinite subset we're talking about. If it's not bounded it's obvious that it has an infinite subset which doesn't have a limit point so it must be bounded.
If it's bounded take $\overline{E}$ so the closure of $E$. This is closed and bounded which is equivalent to compactness in metric spaces. Now if $F \subseteq \overline{E}$ is an infinite subset that has no limit point in $\overline{E}$ then $\forall x \in \overline{E}$ we have a neighborhood such that $\overline{E} \cap N(x)$ either has one element if $x \in F$ or no element if not. But these neighborhoods add an open cover of $\overline{E}$ thus have a finite subcover and at least one of those neighborhoods must have an infinite number of elements which is a contradiction. Note that if $F \subseteq E \subseteq \overline{E}$ then by the previous proof it has a limit point in $\overline{E}$ thus has a limit point but not necessarily in $E$.
