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Presumably, the approach is the one mentioned on [Wikipedia](https://en.wikipedia.org/wiki/Tychonoff%27s_theorem#Proofs_of_Tychonoff's_theorem), specifically the third proof by Cartan and Bourbaki. It is summarized there, *under the assumption, which requires the Ultrafilter Lemma, that a topological space is compact if and only if all ultrafilters on that space converge*.

Given that, the proof is simply we given an ultrafilter on the product consider the ultrafilter generated by the image of a projection. That ultrafilter converges to some point in the corresponding component space, because it's compact. The ultrafilter on the product space then converges to tuple of all these points. Thus, every ultrafilter on the product space converges, and the product space is compact.

This is a close paraphrase of what is stated on Wikipedia, but it is a bit misleading. If you skip to the next paragraph and continue on, you'll see a proof that uses the Ultrafilter Lemma but not the Axiom of Choice. Since the Ultrafilter Lemma is weaker than the Axiom of Choice, but Tychonoff's Theorem is equivalent to it, the Axiom of Choice must be coming in somewhere, and here is where it is. The theorem does not give a unique point for each component. We only end up with a non-empty set. We need the Axiom of Choice to make a tuple where each component chooses an element from the corresponding non-empty set.

With that out of the way, the real substance is thus the claim that a space being compact is equivalent to all ultrafilters converging given the Ultrafilter Lemma.

First, to actually define convergence of (ultra)filters, an (ultra)filter [**converges to a point**](https://en.wikipedia.org/wiki/Filter_(mathematics)#Convergent_filters_and_cluster_points) $x$ if the (ultra)filter contains the neighborhood filter of $x$. The [**neighborhood filter**](https://en.wikipedia.org/wiki/Filter_(mathematics)#Neighborhood_bases) is the set of all subsets that contain an open neighborhood of $x$. Alternatively, it is the upward closure of all open neighborhoods of $x$. This is easily verified to actually be a [filter](https://en.wikipedia.org/wiki/Filter_on_a_set#Definition).

For this, it's best to use the contrapositive [characterization of compactness](https://en.wikipedia.org/wiki/Finite_intersection_property#Compactness):

> A space is **compact** if and only if every family of closed subsets having the finite intersection property has non-empty intersection.

The [**finite intersection property**](https://en.wikipedia.org/wiki/Finite_intersection_property#Definition) simply states that for a collection of sets, any finite subset of them has a non-empty intersection. The important thing is that any non-empty collection of sets with the finite intersection property is a [**filter subbase**](https://en.wikipedia.org/wiki/Ultrafilter_on_a_set#Definitions) from which we can generate a proper filter.

With this, we can prove that compactness implies that every ultrafilter converges. **Proof**: $\implies$ Given an ultrafilter on a compact space, it satisfies the finite intersection property by definition. In particular, if we restrict to the necessarily proper filter generated by the [basis](https://en.wikipedia.org/wiki/Filter_(mathematics)#Filter_bases) of closed subsets contained in the ultrafilter, those also satisfy the finite intersection property, and, thus, by compactness, their intersection is non-empty. Let $x$ be a point in that intersection and let $U$ be any open neighborhood of $x$. Then, by definition, either $U$ is in the ultrafilter or its complement is in the ultrafilter. But its complement is a closed subset that doesn't contain $x$, so it can't be in the ultrafilter. Therefore, every open neighborhood of $x$ is in the ultrafilter and the ultrafilter converges to $x$. $\square$ (Note, this doesn't exclude it from also converging to a different point, e.g. if the points are [topologically indistinguishable](https://en.wikipedia.org/wiki/Topological_indistinguishability).)


Proving the other direction is where the [**Ultrafilter Lemma**](https://en.wikipedia.org/wiki/Ultrafilter_on_a_set#The_ultrafilter_lemma) comes in. It states that every proper filter on a set is contained in some ultrafilter on that set.

With this, it's easy to show that every ultrafilter converging implies compactness. **Proof**: $\impliedby$ Given any collection of closed subsets satisfying the finite intersection property, we get an ultrafilter that converges to some point, $x$, i.e. that ultrafilter contains all the subsets in the neighborhood filter of $x$. If some closed subset in the original collection (and thus in the ultrafilter) *doesn't* contain $x$, then its complement does and is open. It is thus an open neighborhood of $x$ and contained in the ultrafilter. But an ultrafilter can't contain both a set and its complement, so this is a contradiction, so no such closed subset exists. Which is to say, the intersection of *all* the original closed subsets contains $x$. $\square$