Ultrafilter Lemma/Proof 2

Proof
Let $\FF$ be a filter on $S$.

Let the power set of $S$ be ordered by inclusion.

Then, by Filter on Set is Proper Filter, $\FF$ is a proper filter on $\powerset S$.

By Filter is Ideal in Dual Ordered Set, $\FF$ is an ideal in the dual of $\powerset S$.

By Singleton of Bottom is Ideal, $\set \empty$ is an ideal in $\powerset S$.

By Ideal is Filter in Dual Ordered Set, $\set \empty$ is a filter on the dual of $\powerset S$.

By the definition of Filter on Set, $\empty \notin \FF$, so $\FF$ and $\set \empty$ are disjoint.

Therefore, by the Boolean Prime Ideal Theorem, there is a prime ideal $\GG \supseteq \FF$ that is disjoint from $\set \empty$.

By Prime Ideal is Prime Filter in Dual Lattice, $\GG$ is a prime filter on the dual of the dual of $\powerset S$.

Thus, $\GG$ is a prime filter on $\powerset S$ by Dual of Dual Ordering.

But $\GG$ is disjoint from $\set \empty$, which implies that $\empty \notin \GG$.

Therefore, $\GG$ is a proper filter.

Thus, by Proper and Prime iff Ultrafilter in Boolean Lattice, $\GG$ is an ultrafilter on $\powerset S$.

But by comparing the definitions, $\GG$ is thus an ultrafilter on $S$.