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Let $X$ be a topological space and $K$ be a subset of $X$. Let $B$ be a basis of topology on $X$ (i.e. every open subset of $X$ is a union of some members in $B$).

Claim: If every open cover of $K$ by elements of $B$ has a finite sub-cover then $K$ is compact.

Proof: Let $\{ U_{\alpha}:\alpha\in I\}$ be an arbitrary open cover of $K$. We want to find finite sub-cover of it. Now each $U_{\alpha}$ is equal to union of some members of $B$. Thus $K$ is contained in a union of members of $B$ and by hopothesis, it has a finite sub-cover. Thus, if $K\subseteq B_1 \cup B_2 \cup \cdots \cup B_n$, then for each $i$ in $1,2,\ldots, n$, pick-up one member $U_{\alpha_i}$ such that $B_i\subseteq U_{\alpha_i}$. It is then clear that $K$ is covered by $U_{\alpha_1}, U_{\alpha_2},\cdots,U_{\alpha_n}$.

Q.1 Is this argument correct?

Q.2 Using this fact, I think we can easily prove that if $K_1$ and $K_2$ are compact subsets of $X_1$ and $X_2$ then $K_1\times K_2$ is compact subset of $X_1\times X_2$. Is this correct?

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Yes, the proof you provided is exactly correct .

Now, for the $2nd$ problem, this typical fact is not enough as because if $I$ be an index set, then obviously $\cup_{\alpha \in I} (U_{\alpha} × V_{\alpha}) \neq (\cup_{\alpha} U_{\alpha}) × (\cup_{\alpha} V_{\alpha})$; in general, for some family of sets $\{U_{\alpha}|\alpha \in I\}$ and $\{V_{\alpha}|\alpha \in I\}$ .

The crucial fact required here is the somewhat called "The tube lemma".

If $X$ and $Y$ be two topological spaces with $Y$ being compact, then if $S$ be a non-empty subset of $X$ and if $W$ be a neighborhood of $(S×Y)$, then then there exists a point $s \in S$ and a neighborhood $P \subseteq W$ with $s \in P$ such that the set $\{s\} × Y$ is totally contained in $P$ .

Consider any book on basic general topology , rather Munkres and you can find the proof there .

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