Cluster Point of Ultrafilter is Unique

Theorem
Let $S$ be a set.

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

Let $x \in S$ be a cluster point of $\mathcal F$.

Then there is no point $y \in S: y \ne x$ such that $y$ is also a cluster point of $\mathcal F$.

Proof
Let $x, y$ both be cluster points of an ultrafilter $\mathcal F$ on a set $S$.

Then by definition:
 * $\forall U \in \mathcal F: \left\{{x, y}\right\} \subseteq U$

But then we can create a finer filter $\mathcal F \cup \left\{{\left\{{x}\right\}}\right\}$ on $S$.

So $\mathcal F$ is not an ultrafilter.