Munkres' Topology says
Theorem 28.1. Compactness implies limit point compactness, but not conversely.
Proof. Let $X$ be a compact space. Given a subset $A$ of $X$, we wish to prove that if $A$ is infinite, then $A$ has a limit point. We prove the contrapositive - if $A$ has no limit point, then $A$ must be finite. Suppose $A$ has no limit point. Then $A$ contains all its limit points, so that $A$ is closed. ....
That's confusing. Do they mean vacuously $A$ must be closed since it has no limit points even to contain?