How to show the intersection of two compact subsets is compact Let (X,d) be a metric space and A,B $\subset$ X be two compact subsets.
Show $A\cap B$ is also compact.
I attempted this question by showing the intersection is bounded and closed. 
But I stated that Bounded and Closed $\Rightarrow$ Compact (Heine-Borel) but I didn't realise this only holds for $\mathbb R^n$.
Most of the other similar problems on here were dealing with $\mathbb R^n$, so how would you go about showing this for a general space X?
Any help would be greatly appreciated!
 A: In metric space, any compact subset is closed. In particular, this means $X \setminus B$ is open.
For any open cover $\mathscr{O}$ of $A \cap B$, $\mathscr{O} \cup \{ X \setminus B \}$ will be an open cover of $A$. Since $A$ is a compact, $\mathscr{O} \cup \{ X \setminus B \}$ has a finite sub-cover $\mathscr{F}$. It is easy to see $\mathscr{F} \setminus \{ X \setminus B \}$ is a finite sub-cover of $\mathscr{O}$ for $A \cap B$. Since the open cover $\mathscr{O}$ is arbitrary, $A \cap B$ is compact.
A: This is true for arbitrary Hausdorff spaces, not only for metric spaces.
Try to prove the following slight generalisation: any closed set in a compact space is compact. This should be easy with the usual definition of compactness (any cover admits a finite subcover). If you insist on working with metric spaces, it's even easier, using the definition that a set in a metric space is compact iff every sequence has a convergent subsequence.
A: $A$ is closed in $X$, so $A \cap B$ is closed in $B$. Now, a closed subset of a compact space is compact.
