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I want to prove that if $E$ and $F$ are compact sets, then $E\times F$ is compact. Here's my proof:

Let $\{G_\alpha\}$ be an open cover of $E \times F$. For each $(a,b) \in E \times F$, we can choose some $\alpha$ such that $(a,b) \in G_\alpha$. Since $G_\alpha$ is open, the point $(a,b)$ is contained in some open box $U_{(a,b)} \times V_{(a,b)} \subset G_\alpha$, where $U_{(a,b)} \subset E$ and $V_{(a,b)} \subset F$.

Suppose we fix $a$ and vary $b$. Then for every point $(a,b)$ we find that the point is contained in an open box in the product $E \times F$, and that box is then itself the product of a subset of $E$ with a subset of $F$. Proceeding in this manner, we observe that the collection of sets $\{V_{(a,b)}\}_{b\in F}$ is an open cover of $F$. Since by assumption $F$ is compact, we can find a finite cover $\{V_{(a,b_j(a))}\}$ of $F$ that consists of finitely many open sets containing points $\{(a,b_j(a))\}$.

Now let $U_a = \bigcap_j U_{(a,b_j(a))}$. Since $U_a$ is the intersection of finitely many open sets, it is itself open. Since $E$ is compact, there are finitely many $a_i$ such that $\{U_{a_i}\}$ forms an open cover of $E$. Then it follows that the collection of sets $\{U_{a_i}\times V_{(a_i,b_j (a_i))}\}$ (for all $i$ and $j$) is a finite subcover of $E \times F$, hence $E \times F$ is compact.

Since I am not so good at topology yet, I am not entirely sure that this is correct. If is is correct, then there is something I can do to simplify it? If it is wrong, how could I fix it?

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Your proof is correct and this is the standard proof found in most texts, commonly referred to as the "tube lemma".

Perhaps to get more familiarity, look at a nice application: Whitehead's theorem on quotient map. This result is used a lot in algebraic topology.

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