Cluster Point of Ultrafilter is Unique

Theorem
Let $S$ be a set.

Let $\FF$ be an ultrafilter on $S$.

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

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

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

Then by definition:
 * $\forall U \in \FF: \set {x, y} \subseteq U$

But then we can create a finer filter $\FF \cup \set {\set x}$ on $S$.

So $\FF$ is not an ultrafilter.