Since all of the answers to this question(except the one involving Alexander's subbase lemma) refer to a usually strange rehashing of the ultrafilter proof (BOO), I decided to give two nice proofs to Tychonoff's theorem here for Hausdorff spaces.

The first proof of Tychonoff's theorem for Hausdorff spaces uses the Stone-Cech compactification. This proof is useful when one constructs the Stone-Cech compactification before Tychonoff's theorem.

Proof: Assume that $X_{i}$ is compact for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ be the product space. Then each projection $\pi_{i}:X\rightarrow X_{i}$ extends to a continuous map $\overline{\pi_{i}}:\beta X\rightarrow X_{i}$ since each $X_{i}$ is compact. Therefore the map $f:\beta X\rightarrow X$ where $f(x_{i})_{i\in I}=(\overline{\pi_{i}}(x))_{i\in I}$ is a continuous surjection, so $X$ is compact being the continuous surjective image of $\beta X$. QED

For the second proof we use the following facts about uniform spaces that every mathematician should be aware of.

i. Every compact Hausdorff space has a unique compatible uniformity and that uniformity is complete and totally bounded.

ii. If a uniform space is complete and totally bounded, then it is compact.

Tychonoff's theorem then immediately follows from the fact that the product of complete uniform spaces is complete and that the product of totally bounded uniform spaces is totally bounded. And this proof is intuitive because it is easier to imagine that the product of complete and totally bounded uniform spaces is complete and totally bounded than to imagine that the product of compact spaces is compact.

Since all of the answers to this question  (except the one involving Alexander's subbase lemma) refer to a usually strange rehashing of the ultrafilter proof (BOO), I decided to give two nice proofs to Tychonoff's theorem here for Hausdorff spaces.

The first proof of Tychonoff's theorem for Hausdorff spaces uses the Stone-Cech compactification. This proof is useful when one constructs the Stone-Cech compactification before Tychonoff's theorem.

Proof: Assume that $X_{i}$ is compact for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ be the product space. Then each projection $\pi_{i}:X\rightarrow X_{i}$ extends to a continuous map $\overline{\pi_{i}}:\beta X\rightarrow X_{i}$ since each $X_{i}$ is compact. Therefore the map $f:\beta X\rightarrow X$ where $f(x_{i})_{i\in I}=(\overline{\pi_{i}}(x))_{i\in I}$ is a continuous surjection, so $X$ is compact being the continuous surjective image of $\beta X$. QED

For the second proof we use the following facts about uniform spaces that every mathematician should be aware of.

i. Every compact Hausdorff space has a unique compatible uniformity and that uniformity is complete and totally bounded.

ii. If a uniform space is complete and totally bounded, then it is compact.

Tychonoff's theorem then immediately follows from the fact that the product of complete uniform spaces is complete and that the product of totally bounded uniform spaces is totally bounded. And this proof is intuitive because it is easier to imagine that the product of complete and totally bounded uniform spaces is complete and totally bounded than to imagine that the product of compact spaces is compact.