Existence of Nonprincipal Ultrafilter

Theorem
Let $S$ be an infinite set.

Then there exists an nonprincipal ultrafilter $U$ on $S$.

Proof
Let $\FF$ be the Fréchet filter on $S$.

By Fréchet Filter is Filter, $\FF$ is a filter on $S$.

Then, by the Ultrafilter Lemma, there exists some ultrafilter $U \supseteq \FF$ on $S$.

Let $a \in S$ be arbitrary.

By Existence of Singleton Set, there exists a singleton $\set a$.

Let $A = S \setminus \set a$.

By the definition of set difference, $a \notin A$.

Then:

But by Singleton is Finite, $\set a$ is a finite set.

Therefore, $S \setminus A$ is finite as well.

But then, by definition, $A$ is cofinite in $S$.

Thus, by definition of Fréchet filter, $A \in \FF$.

But as $U \supseteq \FF$, $A \in U$ by definition of superset.

Therefore, as $a \in S$ was arbitrary:
 * For every $a \in S$, there is some $A \in U$ such that $a \notin A$.

Thus, $U$ does not have a cluster point, and is nonprincipal by definition.