= Solution
{c}
A set is <compact> if every <open cover> of it has a finite subcover. Let $K_1,\ldots,K_m$ 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 $\bigcup_jK_j$. Hence \b[a finite union of compact sets is compact].
An arbitrary union need not be compact. Each singleton $\{n\}$, $n\geq1$, is compact, but their union $\mathbb N$ has the open cover $\{(n-1/3,n+1/3):n\geq1\}$ with no finite subcover. Every member covers only one positive integer, so finitely many leave infinitely many integers uncovered.
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