By Zorn's lemma, let $M$ be an maximal element in $P=\{G\mid F\subseteq G\}$ where $F$ is a filter on $X$ and $X$ is a non-empty set. Clearly $M$ is also a filter on $X$ by closure under supersets, as $F$ is a filter.
Let $A\subseteq X$. Suppose for a contradiction $A\notin M$ and $X-A\notin M$.
If $A\in F$, then $A\in M$ as $F\subseteq M$. Contradiction.
If $A\notin F$, and $A=\emptyset$, then $X-A \in M$. Contradiction.
But how can I show that in the case of $A\notin F$ and $A\neq\emptyset$, there is a contradiction?
Best Answer
You're approaching this wrong. If $F$ wasn't an ultrafilter, then there will be some $A\in M\setminus F$. So it shouldn't be possible to derive contradiction from the condition $A\notin F$ and $A\neq\varnothing$.
The key point is the following lemma, which you should probably prove if you haven't proved it yet:
Using the lemma we can now appeal to the maximality of $M$ to ensure that for every $A$, either $A\in M$ or $X\setminus A\in M$.