Every Ultrafilter Converges implies Every Filter has Limit Point

Theorem
Let $X$ be a topological space.

Suppose that every ultrafilter on $X$ converges.

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

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

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 $x \in X$.

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

Also See

 * Equivalent Definitions of Compactness