Question seems fine i just have a few doubts.
Is it possible to just use the Heine Borel theorem? as both A and B are compact it implies they are both closed, so therefore their intersection is closed.
I'm just not too sure how to go about showing boundedness. Can you say A ∩ B ⊂ A and A ∩ B ⊂ B where A is bounded above and below by a and b resp. and B is bounded above and below by c and d resp. which implies b is a lower bound and c is an upper bound for A ∩ B ? Therefore it is bounded and compactness follows.
Best Answer
You could not use boundedness to prove compactness in general metric spaces. Take any infinite set $M$ with the metric $d(x,y)=1$ if $x\not=y$. This makes $M$ a metric and a topological space with the discrete topology, every point is an open set (since $B(x,1/2)=\{x\}$ for every $x\in M$). Then the open cover $\{\{x\}: x\in M\}$ has no finite subcover, hence $M$ is not compact. Nevertheless $M\subseteq B(x,2)$ (any $x$), thus $M$ is bounded.
To prove that $A\cap B$ is compact in general (if both are compact subsets of a metric space $X$), take any finite open cover of $A\cap B$ and add to it the set $X\setminus(A\cap B)$ (which is open) so you get a finite cover of $A$ (and also of $B$) so now take a finite subcover of this, using that $A$ is compact. Then throw out the set $X\setminus(A\cap B)$ (if it is in your finite subcover of $A$), what remains is a finite subcover of $A\cap B$.
You didn't ask about it, but you might try to use this idea to prove that $A\cup B$ is also compact.