= Solution
A <filter on a set> $I$ means a proper family $\mathcal F\subseteq\mathcal P(I)$ with $I\in\mathcal F$, $\varnothing\notin\mathcal F$, closure under finite intersections, and upward closure under inclusion. An <ultrafilter> is a maximal proper filter; equivalently, for every $X\subseteq I$, exactly one of $X$ and $I\setminus X$ belongs to it. A <principal ultrafilter> is $\{X\subseteq I:i\in X\}$ for some $i\in I$. A <nonprincipal ultrafilter> is one not of that form, equivalently one containing no finite set.
To prove the <ultrafilter lemma> using <Zorn's lemma>, order all proper filters containing $\mathcal F$ by inclusion. The union of any chain is a proper filter: finitely many members lie together in one filter of the chain, and no chain member contains the empty set. Thus every chain has an upper bound, and Zorn's lemma gives a maximal member $\mathcal U$.
For completeness, its decision property follows from maximality. If $X\notin\mathcal U$, adjoining $X$ cannot generate a proper filter. Hence some $F\in\mathcal U$ has $F\cap X=\varnothing$. Upward closure gives $I\setminus X\in\mathcal U$. Both complements cannot be members, since their intersection is empty. Therefore $\boxed{\mathcal F\subseteq\mathcal U\text{ for an ultrafilter }\mathcal U}$. Properness is essential; a family containing the empty set cannot extend to an <ultrafilter>.
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