Here is mine. It's taken from page 11 of "An Introduction To Abstract Harmonic Analysis", 1953, by Loomis:
https://archive.org/details/introductiontoab031610mbp
https://ia800309.us.archive.org/10/items/introductiontoab031610mbp/introductiontoab031610mbp.pdf
(By the way, I don't know why this book is not more famous.)
To prove that a product $K=\prod K_i$ of compact spaces $K_i$ is compact, let $\mathcal A$ be a set of closed subsets of $K$ having the finite intersection property (FIP) — viz. the intersection of finitely many members of $\mathcal A$ is nonempty —, and show $\bigcap\mathcal A\not=\varnothing$ as follows.
By Zorn's Theorem, $\mathcal A$ is contained into some maximal set $\mathcal B$ of (not necessarily closed) subsets of $K$ having the FIP.
The $\pi_i(B)$, $B\in\mathcal B$, having the FIP and $K_i$ being compact, there is, for each $i$, a point $b_i$ belonging to the closure of $\pi_i(B)$ for all $B$ in $\mathcal B$, where $\pi_i$ is the $i$-th canonical projection. It suffices to check that $\mathcal B$ contains the neighborhoods of $b:=(b_i)$. Indeed, this will imply that the neighborhoods of $b$ intersect all $B$ in $\mathcal B$, hence that $b$ is in the closure of $B$ for all $B$ in $\mathcal B$, and thus in $A$ for all $A$ in $\mathcal A$.
For each $i$ pick a neighborhood $N_i$ of $b_i$ in such a way that $N_i=K_i$ for almost all $i$. In particular the product $N$ of the $N_i$ is a neighborhood of $b$, and it is enough to verify that $N$ is in $\mathcal B$. As $N$ is the intersection of finitely many $\pi_i^{-1}(N_i)$, it even suffices, by maximality of $\mathcal B$, to prove that $\pi_i^{-1}(N_i)$ is in $\mathcal B$.
We have $N_i\cap\pi_i(B)\not=\varnothing$ for all $B$ in $\mathcal B$ (because $b_i$ is in the closure of $\pi_i(B)$), hence $\pi_i^{-1}(N_i)\cap B\not=\varnothing$ for all $B$ in $\mathcal B$, and thus $\pi_i^{-1}(N_i)\in\mathcal B$ (by maximality of $\mathcal B$).
Many people credit the general statement of Tychonoff's Theorem to Cech. But, as pointed out below by KP Hart, Tychonoff's Theorem seems to be entirely due to … Tychonoff. This observation was already made on page 636 of
Chandler, Richard E.; Faulkner, Gary D. Hausdorff compactifications: a retrospective. Handbook of the history of general topology, Vol. 2 (San Antonio, TX, 1993), 631–667, Hist. Topol., 2, Kluwer Acad. Publ., Dordrecht, 1998
The statement is made by Tychonoff on p. 772 of "Ein Fixpunktsatz" (DOI: 10.1007/BF01472256, eudml) where he says that the proof is the same as the one he gave for a product of intervals in "Über die topologische Erweiterung von Räumen" (DOI: 10.1007/BF01782364, eudml).
Screenshot added to answer a comment of ACL:
Best Answer
Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC)
Then one observes that
any image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)
any filter in the product space converges if and only if all its projections converge .
You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?