= Solution
A <subset> $K$ is \b[<compact> if every <open cover> has a <finite subcover>]. Explicitly, if $U_i$ are <open sets> of $X$ and $K\subseteq\bigcup_{i\in I}U_i$, there are finitely many indices $i_1,\ldots,i_m$ such that
$$
K\subseteq U_{i_1}\cup\cdots\cup U_{i_m}.
$$
Equivalently, the <subspace topology> makes $K$ a <compact space>. An <open set> in this <subspace topology> is $K\cap U$ for an ambient <open set> $U$, which proves equivalence of the two <open cover> formulations. The <empty set> is <compact>: the empty selection is a <finite subcover>.
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