Cluster Point of Ultrafilter is Unique

Theorem
Let $X$ be a set and let $\mathcal F$ be an ultrafilter on $X$.

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

Then there is no point $y \ne x$ 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 $x$.

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 $X$.

So $\mathcal F$ is not an ultrafilter.