Question about product of compact spaces being compact I'm reading through a proof that for $X,Y$ compact, $X\times Y$ compact with the standard topology.
The proof begins by saying: consider a specific type of cover $\mathcal{C}$ where each $T\subseteq \mathcal{C}$ is of the form $$T=\{U\times V:\;U\in\tau_X, V\in\tau_Y\}$$
The proof is then standard: fix $x\in X$ then $V_x=\{V_{xy}:y\in Y\}$ covers $Y$ and has finite subcover etc. we eventually finitely cover $U_x\times V$, and similarly $X$ has finite cover etc. so we end up with a finite subcover of $\mathcal{C}$.
My question is, isn't the product topology generated by $\{U\times V:\;U\in \tau_X,V\in\tau_Y\},$ so any cover of $X\times Y$ must have the form above anyway? I am a little confused as to how said cover is of a specific type.
 A: That proof relies on a lemma.
Lemma. Suppose $\mathscr{B}$ is a basis for the topology $\tau$ on $X$. Then $X$ is compact if and only if every open cover consisting of elements of $\mathscr{B}$ has a finite subcover.
A basis for $\tau$ is a subset $\mathscr{B}$ of $\tau$ such that every open set $U\in\tau$ is the union of elements in $\mathscr{B}$.
One direction of the lemma is obvious. So suppose $\mathscr{B}$-covers (covers consisting of elements of $\mathscr{B}$) have finite subcovers. Let $\mathscr{U}=\{U_\alpha: \alpha\in A\}$ be any cover of $X$.
For each $x\in X$ there exists $\alpha(x)$ such that $x\in U_{\alpha(x)}$ and also $V_x\in\mathscr{B}$ such that $x\in V_x\subseteq U_{\alpha(x)}$, since $\mathscr{B}$ is a basis. The set $\{V_x:x\in X\}$ is a $\mathscr{B}$-cover of $X$, so it has a finite subcover, say
$$
X=V_{x_1}\cup V_{x_2}\cup\dots\cup V_{x_n}
$$
Since
$$
V_{x_1}\cup V_{x_2}\cup\dots\cup V_{x_n}\subseteq
U_{\alpha(x_1)}\cup U_{\alpha(x_2)}\cup \dots\cup U_{\alpha(x_n)}
$$
we have that $\{U_{\alpha(x_1)}, U_{\alpha(x_2)}, \dots, U_{\alpha(x_n)}\}$ is a finite subcover of $\mathscr{U}$. QED
In the case of the product topology, the sets of the form $U\times V$, for $U\in\tau_X$ and $V\in\tau_Y$ form a basis for the product topology, so you can prove compactness using just those special covers.
Note: a similar result holds for subbases, but it's harder to prove. See, for example, Kelley's book.
A: Consider say $[0,1] \times [0,1].$ There are open sets in the product which are not products of one open set from each factor, e.g. the interior of a circle contained in that square. I think by saying the open sets are generated by those you mention one is working with a subbasis for the proof, which seems like it is OK to do, but not directly without some argument.
Edit: I used the wrong term "subbasis" here where it should be "basis" as @egreg points out in his (rather complete) answer.
