Proof Let $(X_j,d_j)$ be a compact metric space for $j=1,...,n$. Denote $X=X_1 \times X_2 \times...\times X_n$ to be the product of compact metric spaces.
Assume the property that the open sets in $(X,d)$ are unions of product sets of the form $U_1 \times ... \times U_n$, where $U_j$ is an open subset of $X_j$.
Suppose $X\subset \bigcup_{\alpha\in A} U^{\alpha}$, where $U^{\alpha}=U_1^{\alpha}\times ... \times U_n^{\alpha}$, and $U_j^{\alpha}\subset X_j$ are open.
Now, we have that $X_j\subset \bigcup_{\alpha\in A} U_j^{\alpha}$, for if $x_j\in X_j$, $\exists y\in X$ such that $y_j=x_j$, but $y\in U^{\alpha}=U_1^{\alpha}\times ... \times U_n^{\alpha}$ for some $\alpha$.
$\implies y_j=x_j\in U_j^{\alpha}$.
So by compactness of $X_j$, $\exists$ a finite subcover $X_j\subset \bigcup^{m(j)}_{i=1} U_j^i$, where $X_j$ is covered by $m(j)$ open sets for each $j=1,...,n$.
Finally, $X\subset \bigcup_{1\leq i_j\leq m(j)} (U_1^{i_1} \times U_2^{i_2} \times ...\times U_n^{i_n})$ is a union of $\Pi^n_{i=1}m(i)<\infty$ open subsets of $X$, hence $X$ is compact. end Proof
Is this a valid way of arguing? My main concern is that the open cover at the end of the proof is not actually a subcover... or that it contains a subcover plus some surplus open sets.