If A and B are compact sets, then A ∩ B is compact.
my approach is since A and B are compact, for every open covers of A and B, there exists finite sub-covers to cover A and B. Since A ∩ B is a subset of A, for each finite sub-cover of A, there is a sub-cover of that sub-cover to cover A ∩ B. Therefore, A ∩ B is compact. Is my proof rigorous enough?
You have to assume that the space is Hausdorff so that compact sets ar closed. $A\subset \cup_i V_i$ ($V_i$'s open ) implies $A \subset B^{c} \cap \cup_i V_i$ and this gives an open cover for $A$. Let $A \subset B^{c} \cap \cup_{k=1}^{N} V_{i_k}$. Then $A \cap B$ is covered by $V_{i_k}$,$1\leq k\leq N$.