Prove that the intersection of two compact sets is compact, using the Heine-Borel Criterion Using the fact that:

S is compact: every open cover has a finite subcover.

Prove:
Given $A,B \subset \Bbb R^n$ are compact sets, then $A\cap B$ is compact.

Here is my attempt, using the fact that $A$ follows the above Heine-Borel criterion above and $A \cap B \subset A$: 
Since $A$ is compact, every open cover of $A$ has a finite subcover. We wish to show that every open cover in $A \cap B$ has a finite  subcover.
Since $A \cap B \subset A$, then every open cover of A must be an open cover of $A \cap B$ (from the definition of an open cover shown below).

A collection of sets ${U_\alpha}$ is an open cover os $S$ if $S$ is contained in $\bigcup U_\alpha$.

Since $A$ is compact, we know that every open cover has a finite subcover. 
Therefore, since $A \cap B \subset A$ and $A$ has a finite subcover for every open cover, $A \cap B$ has a finite subcover for every open cover. 
Is this the correct way to approach this problem? 
Thanks!
 A: A red flag here is that you never used the fact that $B$ is compact. And you have to be more careful about what exactly you mean by a "cover." 
You are given that $A$ and $B$ are compact subsets of $\mathbb R^n$ and you want to prove that $A\cap B$ is compact, using only the definition. So, give $A\cap B$ the subspace topology, and let $\{U_{\alpha}\}_{\alpha\in \Lambda}$ be an open cover of $A\cap B$. By definition of the subspace topology, there are opens $\{O_{\alpha}\}_{\alpha\in \Lambda}$ in $\mathbb R^n$ such that $O_{\alpha}\cap(A\cap B)=U_{\alpha}$. Then, and here we use the fact that $A\cap B$ is closed (because it is the intersection of compact, hence closed sets), the sets $\{O_{\alpha}\}_{\alpha\in \Lambda}\cup \mathbb R^n\setminus A\cap B$ form an open cover of $A$, and now we can extract a finite subcover which covers $A$, hence covers $A\cap B$. And since $\mathbb R^n\setminus A\cap B$ is $\textit{not}$ a covering element of $A\cap B$ so it must be that the finite cover consists of elements of $\{O_{\alpha}\}_{\alpha\in \Lambda}$ alone. And now, unwinding the definitions, we get a finite subcover of the original cover $\{U_{\alpha}\}_{\alpha\in \Lambda}$.
A: In a very general approach you can consider the following result:

Proposition: Let $(X, \mathscr{T})$ be an arbitrary topological space, $K \subseteq X$ be compact (with the relative topology) and $F \subseteq X$ a closed subset. Then $K\cap F$ is also compact. 

Proof: As the intersection with an absolutely closed subset, $K \cap F$ will be closed relatively to the subspace topology on $K$; the result immediately follows from the general proposition that closed subsets of compact spaces are compact. $\Box$
This general result applies indeed to your case since any compact subset in a Hausdorff space is necessarily closed.
Try to see if you can understand and prove the proposition mentioned at the end, let me know if you need more details.  
