Every Ultrafilter Converges implies Every Filter has Limit Point

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

Let every ultrafilter on $S$ be convergent.

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 hypothesis, $\mathcal F'$ converges to some $x \in S$.

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

Also see

 * Equivalence of Definitions of Compact Topological Space