Vtyjkyuk

I want to prove that if $E$ and $F$ are compact sets, then $Etimes F$ 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 $V_(a,b)_bin F$ 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 $U_a_itimes V_(a_i,b_j (a_i))$ (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?

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.