Every Ultrafilter Converges implies Every Filter has Limit Point

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Suppose that every ultrafilter on $S$ converges.

Then every filter on $S$ has a limit point.

Proof
Let $\mathcal F$ be a filter on $S$.

By the Ultrafilter Lemma, there exists an ultrafilter $\mathcal F'$ such that $\mathcal F \subseteq \mathcal F'$.

By $(4)$ we know that $\mathcal F'$ converges to some $x \in S$.

Thus by Limit Point iff Superfilter Converges implies that $x$ is a limit point of $\mathcal F$.

Also see

 * Equivalent Definitions of Compactness