Proof:
I am following this. But, I feel, I am missing something.
Consider two compact spaces $X_1$ and $X_2$ and some cover $U$ of their product space. Consider an element $x\in X_1$. The sets $A_{x,y}$ within $U$ contain $(x,y)$ for each $y\in X_2$.
Now, define $\pi_2:A_{(x,y)}$ which forms an open cover for $X_2$ and has finite subcover $\{A_{(x,y_i)}\}$.
It's clear up to this point.
Let $\{B_x\}$ be the intersection of $\{\pi_1:(A_y)\}$ within $A_{y_i}$, which is open. Thus, $\{B_x\}$ forms an open cover, which has a finite subcover, $\{B_{x_i}\}$. The corresponding sets $\{A_{x_i,y_i}\}$ is finite, and forms an open subcover of the set.
First, we know the covers cover $X_1$ and $X_2$ respectively. But, how do we know that they cover $X_1*X_2$? Second, I believe $\pi_1$ is not necessarily non-empty. So how can we be sure that it is open?
I would appreciate any comments or other opinions.
P.S. The notations may be, like in original source, somewhat ambigious too.
Best Answer
Each $x\in X$ is element of some $B_{x_j}=\bigcap_{i=1}^{n_{x_j}}\pi_X(A_{x_j,y^{x_j}_i})$. It suffices to show that $\{x\}\times Y\subseteq\bigcup_{i=1}^{n_{x_j}}A_{x_j,y^{x_j}_i}$. (I put some super-index on the $y_i$'s because they depend on $x_j$.) At this point it is crucial that the $A\in S$ are products of open sets in $X$ and $Y$, and not just open sets in $X\times Y$. Then, since $x\in\bigcap_{i=1}^{n_{x_j}}\pi_X(A_{x_j,y^{x_j}_i})$ and for each $y\in Y$ there is a $y^{x_j}_k$ ($1\le k\le n_{x_j}$) such that $y\in\pi_Y(A_{x_j,y^{x_j}_k})$, you can derive that $(x,y)\in A_{x_j,y^{x_j}_k}$.