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 $\mathcal B$).
A pdf version is available at http://iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Tycho/ .
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. 272 of "Ein Fixpunktsatz"
http://www.springerlink.com/content/n61706447r886l58/?p=328f0106a2634abfb53531fca0ca5a90&pi=0
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"
http://www.springerlink.com/content/l656352441w67612/?p=328f0106a2634abfb53531fca0ca5a90&pi=1
