Proof of Tychonoff's Theorem
Theorem
Suppose an index set $\mathscr{A}$ is given.
If $\left\{ X_{\alpha} \ | \ \alpha \in \mathscr{A} \right\}$ is a set of compact spaces, then $\displaystyle X : = \prod_{\alpha \in \mathscr{A}} X_{ \alpha}$ is compact.
Explanation
For a theorem important enough to bear a name, it looks like just some trivial property, but in fact the opposite view is correct: it looks like a trivial property, yet it turns out to be so surprisingly hard to prove that the theorem ended up with a name attached to it. That compactness, second to none in usefulness, is preserved even under the Cartesian product of topological spaces is as good as it gets.
Proof
Let $U_{\alpha} \subset X_{\alpha}$ be an open set of $X_{\alpha}$, and for the projection $p_{\alpha} : X \to X_{\alpha}$, define a subbasis generating the product topology of $X$ as follows. $$ \mathscr{S} : = \left\{ p_{\alpha}^{-1} ( U_{\alpha} ) \ | \ U_{\alpha} \subset X_{\alpha} , \alpha \in \mathscr{A} \right\} $$ To use the Alexander subbase theorem, we will show that every open cover from $\mathscr{S}$ has a finite subcover.
Alexander subbase theorem: Let $X$ be a topological space. $X$ is compact. $\iff$ There exists some subbasis $\mathscr{S}$ of $X$ such that every open cover of $X$ consisting of members of $\mathscr{S}$ has a finite subcover.
Part 1.
Assume that there exists an open cover $\mathscr{U} \subset \mathscr{S}$ that has no finite subcover.
Let $V$ denote an open set in $X_{\alpha}$. For every $\alpha \in \mathscr{A}$, if we set $$ \mathscr{U}_{\alpha} = \left\{ V \mid p_{\alpha}^{-1} (V) \in \mathscr{U} \right\} $$ then $\mathscr{U}$ is as follows. $$ \mathscr{U} = \bigcup_{\alpha \in \mathscr{A}} \left\{ p_{\alpha}^{-1} (V) \ | \ V \in \mathscr{U}_{\alpha} \right\} $$
Part 2. For $\alpha \in \mathscr{A}$, we show that $\mathscr{U}_{\alpha}$ is not an open cover of $X_{\alpha}$.
Assume that $\mathscr{U}_{\alpha}$ covers $X_{\alpha}$.
Since $X_{\alpha}$ is compact, there exist $U_{\alpha_{1}} , \cdots , U_{\alpha_{n}} \in \mathscr{U}_{\alpha}$ satisfying $\displaystyle X_{\alpha } = \bigcup_{i=1}^{n} U_{\alpha_{i}}$. Then $$ \left\{ p_{\alpha}^{-1} ( U_{\alpha_{i}} ) \ | \ i = 1, \cdots , n \right\} $$ becomes a finite subcover of $\mathscr{U}$, which contradicts the definition of $\mathscr{U}$ in Part 1. Therefore, $\mathscr{U}_{\alpha}$ fails to be a cover of $X_{\alpha}$.
Part 3.
Since $\mathscr{U}_{\alpha}$ is not a cover of $X_{\alpha}$, for every $U_{\alpha} \in \mathscr{U}_{\alpha}$ there exists an $x_{\alpha} \in X_{\alpha}$ satisfying $x_{\alpha} \notin U_{\alpha}$. Then for every $U \in \mathscr{U}$, we have $p_{\alpha}^{-1} (x_{\alpha} ) \notin U$, and $\mathscr{U}$ fails to be a cover of $X$. This means that there is no open cover $\mathscr{U} \subset \mathscr{S}$ without a finite subcover. By the Alexander subbase theorem, $X$ is compact.
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