What is the importance of Zorn's lemma in Tychonoff theorem proof? 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 $\mathcal{F}$?
 A: So the step you are missing is that the proof you have, very lightly concludes that $\bigcap\overline{H}\neq\emptyset$. Which should be understood as $x\in\bigcap\overline{H}$. But why? This is unclear and to prove that you need $\mathcal{H}$ to be maximal.
First I will show you an example when it fails if $\mathcal{H}$ is not maximal. Consider $X_1=X_2=\{0,1\}$ with discrete topology. Then $X=X_1\times X_2$ has four elements. Let $\mathcal{H}=\{K\}$ consist of a single closed subset $K=\{(0,0),(1,1)\}$. Obviously $\mathcal{H}$ has FIP (Finite Intersection Property). But $\mathcal{H}$ is not maximal since we can extend it with for example $\{(1,1)\}$ and maintain FIP.
Now lets construct $x$. Apply projection $\pi_1$ to $\mathcal{H}$ to get $\pi_1(K)=\{0,1\}$. So we have two choices for $x_1$. Since the choice is arbitrary then pick $x_1=0$. Now apply $\pi_2$ to $\mathcal{H}$ to get $\pi_2(K)=\{0,1\}$ as well. And again the choice of $x_2$ is arbitrary, so lets take $x_2=1$ this time. So we end up with $x=(0,1)$ which doesn't belong to $K$ and therefore it doesn't belong to $\bigcap_{H\in\mathcal{H}}\overline{H}$.
One way to fix that situation is too somehow pick $x_\alpha$ in a smart way. But there doesn't seem to be a way to do that in general. It turns out that it is easier to enlarge $\mathcal{H}$ as much as we can so that any choice of $x_\alpha$ becomes valid.
So as you can see the assumption about $\mathcal{H}$ being maximal is crucial. That's why we can't apply the reasoning to $\mathcal{F}$, because it might not be big enough. Actually $\mathcal{F}$ hardly ever is maximal because it contains closed subsets only.
Anyway the full proof is a bit too long for Math StackExchange so please have a look at details here:
https://www.math.arizona.edu/files/grad/workshops/integration/projects/tychonoff.pdf
Notice that the statement "$x\in\overline{H}$ for each $H$" is not trivial at all. And I'm surprised that whoever wrote the proof you have, treated it as obvious.
