Every Filter has Limit Point implies Every Ultrafilter Converges

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

Suppose that each filter on $S$ has a limit point in $S$.

Then each ultrafilter on $S$ converges to a point in $S$.

Proof
Let $\mathcal F$ be an ultrafilter on $S$.

By $(3)$ $\mathcal F$ has a limit point $x \in S$.

By Limit Point iff Superfilter Converges, there exists a filter $\mathcal F'$ on $S$ which converges to $x$ satisfying $\mathcal F \subseteq \mathcal F'$.

Because $\mathcal F$ is an ultrafilter, $\mathcal F = \mathcal F'$.

Thus $\mathcal F$ converges to $x$.

Also see

 * Equivalent Definitions of Compactness