It is indeed enough to consider basic open covers as you did (despite some comments to the contrary): if $\mathcal{B}$ is a base for a space $(X, \mathcal{T})$ then $X$ is compact iff every cover by elements from $\mathcal{B}$ has a finite subcover:
To see the sufficiency, let $\{O_i: i \in I\}$ be any open cover of $X$.
Then, for each $x$, $x \in O_{i(x)}$ for some $i(x) \in I$, and then there is a basic element $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq O_{i(x)}$.
Then $\{B_x: x \in X\}$ is an open cover of $X$ by members of $\mathcal{B}$, so by assumption this has a finite subcover $\{B_{x_1},\ldots, B_{x_n}\}$.
But then $\{O_{i(x_1)}, \ldots , O_{i(x_n}\}$ is a finite subcover of our original cover (as it is an enlargement of the subcover of base elements). So $X$ is compact.
But taking individual subcovers in the coordinates does not necesarily work, suppose that we were ding $[0,1] \times [0,1]$ and we have $U=[0,\frac{3}{4}) \times U_2$ and $V=(\frac{1}{4},1] \times V_2$ in our basis covers we could pick
the first components as the finite subcover in the first coordinate, but there is no guarantee that $U$ and $V$ then cover $[0,1]\times [0,1]$, because we pick the subcovers in the coordinates independently. This cannot easily be fixed, and you need a stronger idea. The tube lemma is one way to go. Another the Alexander subase theorem, which proves the same fact for subbases as we did for bases above, and is my preferred way (it also gives arbitary products in one fell swoop). Using Alexander's subbase lemma,see Wikipedia as well:
Just note that $$\mathcal{S} = \{\pi_Y^{-1}[V] = X \times V: V \text{ open in } Y\}$ \cup \{\pi_X^{-1}[U] = U \times Y: U \text{ open in } X\}$$ together forms a subbase for $X \times Y$ (their finite intersections are just the standard basic elements $U \times V$, $U,V$ open). Now suppose we have a cover of $X \times Y$ by
the sets $\mathcal{O}=\{\pi_X^{-1}[U_i], \pi_Y^{-1}[V_j], i \in I, j \in J\}$ from $\mathcal{S}$. Suppose the $\{U_i, i \in I\}$ do not form a cover of $X$, so some $x_0 \in X$ does not lie in any $U_i$ and also that $\{V_j : j \in J\}$ is not a cover of $Y$, so some $y_0$ is not covered them. Then clearly $(x_0, y_0)$ is not covered by $\mathcal{O}$ at all, and this cannot be. So one of these families is a cover for $X$ resp. $Y$ and then we apply the compactness to get a finite subcover in that factor and the corresponding inverse images are then the required subcoevr for $\mathcal{O}$. QED.
Note that using the subbase removes the dependency on the other coordinates when choosing a subcover in a coordinate. This proof also works for any product (in he product topolgoy) where a similar natural subbase exists.
If $\langle x,y\rangle\in U\times V$, then $\big(\{x\}\times B\big)\cup\big(A\times\{y\}\big)\subseteq W$, but that’s not enough to ensure that $\langle x,y\rangle\in W$. It would be better to handle one factor at a time. For each $\langle x,y\rangle\in A\times B$ there are open sets $U(x,y)$ in $X$ and $V(x,y)$ in $Y$ such that $\langle x,y\rangle\in U(x,y)\times V(x,y)\subseteq W$. Fix $a\in A$; then $\{V(a,y):y\in B\}$ is an open cover of $B$ in $Y$, so there is a finite $B_a\subseteq B$ such that $\{V(a,y):y\in B_a\}$ covers $b$. Now do what you did in proving that your $U$ is open: let $$U_a=\bigcap_{y\in B_a}U(a,y)\,.$$ Since $B_a$ is finite, this is an open nbhd of $a$ in $X$. Let
$$V_a=\bigcup_{y\in B_a}V(a,y)$$
and
$$G_a=\bigcup_{y\in B_a}\big(U_a\times V(a,y)\big)=U_a\times V_a\,;$$
then $G_a$ is open in $X\times Y$, and $\{a\}\times B\subseteq G_a\subseteq W$.
Now use the compactness of $A$: there is a finite $A_0\subseteq A$ such that $\{U_a:a\in A_0\}$ is an open cover of $A$, so that $A\times B\subseteq\bigcup\{G_a:a\in A_0\}\subseteq W$, and you’re almost done.
Best Answer
Your proof is correct and I do not see that there is shorter way to prove it. Perhaps you should explicitly mention that the closure of the union of finitely many sets agrees with the union of their closures.