 Solution

A set is compact if every open cover of it has a finite subcover. Let $K_1,\dots ,K_m$ be compact and let $\mathcal{U}$ cover their union. For each $j$, the same family covers $K_j$, so compactness supplies a finite subfamily ${\mathcal{U}}_j$ covering $K_j$. The union of these finitely many finite subfamilies is finite and covers ${\bigcup }_j{K}_j$. Hence a finite union of compact sets is compact.

An arbitrary union need not be compact. Each singleton $\{n\}$, $n\ge 1$, is compact, but their union $\mathbb{N}$ has the open cover $\{(n-1/3,n+1/3):n\ge 1\}$ with no finite subcover. Every member covers only one positive integer, so finitely many leave infinitely many integers uncovered.