I read that Tychonoff's theorem states that given a family of topological spaces $\{(X_i,\tau_i),i \in I\}$, the product topology $(X,\tau)=\prod_{i \in I}(X_i,\tau_i)$ is compact iff each $(X_i,\tau_i)$ is compact.
The proof states that given any family $\mathcal{F}$ of closed subsets of $X$ with the finite intersection property, there is a maximal family $\mathcal{H}$ that contains $\mathcal{F}$ and which also has the finite intersection property. The existence of $\mathcal{H}$ is shown using Zorn's lemma.
The proof goes on to show that $\bigcap_{H \in \mathcal{H}}\bar{H} \neq \emptyset \implies \bigcap_{F \in \mathcal{F}}F \neq \emptyset$, from which the theorem follows. I won't go into the details, but the proof took a 'slice' from each $H \in \mathcal{H}$, using a projection function $p_i$, so that $p_i(H) \in X_i$. Since each $p_i(H)$ has the finite intersection property for $i \in I$, and $X_i$ is compact, we can get the intersection $x_i$ of all $p_i(H)$, and form $x=\prod_{i\in I}x_i \in X$. It follows that with $x$, we have $\bigcap_{H \in \mathcal{H}}\bar{H} \neq \emptyset$
I think am missing a step, but what is the importance of having to use Zorn's lemma to get a maximal $\mathcal{H}$ and show that $\bigcap_{H \in \mathcal{H}}\bar{H} \neq \emptyset$. Since $\mathcal{F}$ already has the finite intersection property, can we not use any arbitrary $\mathcal{F}$?