= Solution
Assume $K$ is <compact>, and let the <closed subsets> $F_i$ of $K$ have the <finite intersection property>. If their total <intersection> were empty, their <complements> $K\setminus F_i$ would form an <open cover> of $K$. A <finite subcover> would give
$$
K=\bigcup_{j=1}^m(K\setminus F_{i_j})
\quad\Longrightarrow\quad
\bigcap_{j=1}^mF_{i_j}=\varnothing,
$$
contradicting the <finite intersection property>. Therefore the total <intersection> is nonempty.
Conversely, assume the stated nonempty-<intersection> property for every family of <closed subsets> of $K$. Given an <open cover> $\{V_i:i\in I\}$ of $K$ in its <subspace topology>, set $F_i=K\setminus V_i$. Their total <intersection> is empty. By the contrapositive of the assumed property, this family fails the <finite intersection property>; hence some finite $J\subseteq I$ satisfies
$$
\bigcap_{j\in J}F_j=\varnothing
\quad\Longrightarrow\quad
K=\bigcup_{j\in J}V_j.
$$
This is a <finite subcover>, proving <compactness>. Both directions are instances of <De Morgan's laws>. Thus \b[<compactness> is equivalent to the <finite intersection characterization of compactness>].
For completeness, the <intersection> of the empty family of <subsets> of $K$ is $K$. If $K$ is empty, no family has the <finite intersection property> when the empty finite subfamily is included, so the implication remains valid.
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