 Solution

For each $b\in B$, use the supplied Hausdorff condition fact to choose disjoint open sets $U_b,V_b$ with $A\subseteq U_b$ and $b\in V_b$. Compactness of $B$ gives finitely many $b_1,\dots ,b_r$ whose $V_{b_j}$ cover $B$. SetBoth are open; $U$ contains $A$ and $V$ contains $B$. They are disjoint because membership in $U\cap V$ would put a point in some $U_{b_j}\cap V_{b_j}$. If one compact set is empty, use the empty set and $X$ directly. Thus disjoint compact subspaces of a Hausdorff space admit disjoint open neighborhoods.