A space $S$ is sequentially compact if every sequence has a convergent subsequence. A space $S$ is limit point compact is every infinite subset has a limit point in $S$.
Proving sequential compactness implies limit point compactness more or less involves letting $E \subset S$ be infinite and extracting a countably infinite subset from $E$, which we can then view as a sequence. What the convergent subsequence converges to will be the limit point.
However, the claim that an arbitrary infinite set has a countably infinite subset is dependent on the axiom of choice. Does sequentially compact $\to $ limit point compact still hold without choice? Can my proof above be modified so that choice isn't needed?