The non-standard analysis proof is an interesting "application" of the ultrafilter proof: a topological space $A$ is compact if and only if every point in the associated "non-standard topological space" ${}^*A$ is near-standard, that is to say, if and only if each $x \in {}^*A$ is contained in every open neighborhood of some standard point $y \in A$ ($i.e.,$ for all $U \subset A$, $U$ open and $y \in U$ implies $x \in {}^*U \subset {}^*A$).

So let $\mathcal{X}$ be a set of topological spaces indexed by $I$, and $P$, the product of these spaces; write ${}^\*P$ for the "non-standard product" of the set ${}^*\mathcal{X}$ of topological spaces indexed by ${}^*I$, and let $x \in {}^\*P$. It suffices to show that $x$ is near-standard.

For each $\kappa \in I$, let $x_\kappa \in {}^\*X_\kappa \in {}^*\mathcal{X}$ be the $\kappa$th factor of $x$. Then $x_\kappa$ is necessarily near-standard, because $X_\kappa \in \mathcal{X}$ is compact. But this means we can find a point $y \in P$ with factors $y_\kappa \in X_\kappa$ such that $U \subset X_\kappa$ open and $y_\kappa \in U$ implies $x_\kappa \in {}^*U \subset {}^*X_\kappa$, thus $V \subset P$ open and $y \in V$ implies $x \in {}^*V \subset {}^*P$. But this means $x$ is near-standard, so $P$ is compact.

"Under the hood," this is basically the ultrafilter proof (my favorite, to answer the original question), so the axiom of choice is required in more or less the same places: while the non-standard objects exist by the Boolean prime ideal theorem, "finding" the $y_\kappa$ in non-Hausdorff spaces requires the full axiom of choice.