$\prod X_{a\in A} $ with the box topology is compact show that $X_a$ is compact for every $a\in A$
I tried to take some open cover $\cup U_{b\in B}$ that having finite subcover $U=\cup_{i=1}^N U_i$ = $\prod X _{a\in A}$ does saying that $U\cap X_a$ must be a finite open cover to $X_a$ is enough? because ($U\cap X_a) \subset U$ and U is finite open cover?
Each $X_a$ is the continuous image of the whole product under the $a$-th projection. Projections are continuous (also in the box topology).
Or using open covers: suppose $\mathcal{U}$ is an open cover of $X_b$ for some $b \in A$. Then define for each $O \in \mathcal{U}$ the box open set $b(O):=\prod_a O_a$ with $O_a =X_a$ if $a \neq b$ and $O_b =O$. Then $\{b(O): O \in \mathcal{U}\}$ is an open cover of $\prod_a X_a$, so has a finite subcover $\{b(O_1),\ldots, b(O_n)\}$. These $O_i$ then form the required finite subcover for $\mathcal{U}$ (this is just the continuous projection proof in another guise, really.)
But harder fact: $\prod_{a \in A} X_a$ is compact in the box topology, essentially only if we have a finite product of compact $X_a$. That is, if we assume all $X_a$ are Hausdorff spaces with more than one point, $\prod_{a \in A} X_a$ is not compact in the box topology when $A$ is infinite.