Let $\mathfrak U$ be the family of filters over $X$ which are finer than $\mathcal F$ , under the partial order of inclusion.

Claim 1. Every chain in $\mathfrak U$ has an upper bound also in $\mathfrak U$ .

Proof. Take any chain $\mathfrak C$ in $\mathfrak U$ , and consider the set $\mathcal C = \cup \mathfrak C$ . Then $\mathcal C$ is also a filter: it cannot contain the empty set, since no filter in the chain does; the intersection of two sets in $\mathcal C$ must be present in the filters of $\mathfrak C$ ; and $\mathcal C$ is closed under supersets because every filter in $\mathfrak C$ is. Obviously $\mathcal C$ is finer than $\mathcal F$ .

So we conclude, by Zorn's lemma, that $\mathfrak U$ must have a maximal filter say $\mathcal U$ , which must contain $\mathcal F$ . All we need to show is that $\mathcal U$ is an ultrafilter. Now, for any filter $\mathcal U$ , and any set $Y \subseteq X$ , we must have:

Claim 2. Either $\mathcal U_1 = \{ Z \cap Y : Z \in \mathcal U \}$ or $\mathcal U_2 = \{ Z \cap (X \backslash Y) : Z \in \mathcal U \}$ (or both) are a filter subbasis.

Proof. We prove by contradiction that at least one of $\mathcal U_1$ or $\mathcal U_2$ must have the finite intersection property. If neither has the finite intersection property, then for some $Z_1, \ldots Z_k$ we must have $$ \varnothing = \bigcap_{1\le i \le k} Z_i \cap Y = \bigcap_{1\le i \le k} Z_i \cap (X \backslash Y). $$ But then $$ \varnothing = \left(\bigcap_{1\le i \le k} Z_i \cap Y\right) \cup \left(\bigcap_{1\le i \le k} Z_i \cap (X \backslash Y)\right) = \bigcap_{1\le i \le k} Z_i, $$ and so $\mathcal U$ does not have the finite intersection property either. This cannot be, since $\mathcal U$ is a filter.

Now, by Claim 2, if $\mathcal U$ were not an ultrafilter, i.e., if for some $Y$ subset of $X$ we would have neither $Y$ nor $X \backslash Y$ in $\mathcal U$ , then the filter generated $\mathcal U_1$ or $\mathcal U_2$ would be finer than $\mathcal U$ , and then $\mathcal U$ would not be maximal.

So $\mathcal U$ is an ultrafilter containing $\mathcal F$ , as intended.