If every infinte subset of E has a limit point in E then E is compact in X Let X be a metric space. Show that a subset E of X is compact if and
only if every infinite subset of E has a limit point in E.
How to prove the converse statement ?
i.e. if every infinite subset of E has a limit point in E then E is compact in X.
I want to prove from the definition of compact sets.
First note that by this answer we know that if $X$ limit point compact (i.e. every infinite set has a limit point), then $X$ is Lindelöf (i.e. every open cover has a countable subcover), otherwise $X$ would have a closed and discrete uncountable subspace, contradicting the limit point compactness.

Then as metric spaces are $T_1$, this answer shows that a limit point compact metric $X$ is also countably compact (every countable cover has a finite subcover).

And clearly then $X$ is compact (we reduce an open cover to a countable one via Lindelöfness, and then to a finite one using countable compactness). Also see this answer for an overview of similar ideas.
First note that by this answer we know that if $X$ limit point compact (i.e. every infinite set has a limit point), then $X$ is Lindelöf (i.e. every open cover has a countable subcover), otherwise $X$ would have a closed and discrete uncountable subspace, contradicting the limit point compactness.

Then as metric spaces are $T_1$, this answer shows that a limit point compact metric $X$ is also countably compact (every countable cover has a finite subcover).

And clearly then $X$ is compact (we reduce an open cover to a countable one via Lindelöfness, and then to a finite one using countable compactness). Also see this answer for an overview of similar ideas.