Lemma
A topological space is compact iff each family of closed sets which have the finite intersection property has a non-void intersection.
I've proved the same result in another way but I really can't catch his argument in the second statement of the proof.
Proof
If A is a family of subsets of a topological space X, then, according to the De Morgan formulae, $X\setminus\bigcup\{a : a\in A\} = \bigcap\{X\setminus a : a\in A\}$ and hence A is a cover of X iff the intersection of the complements of the members of A is void. The space X is compact iff each family of open sets, such that no finite subfamily covers X, fails to be a cover, and this is true iff each family of closed sets which possesses the finite intersection property has a non-void intersection.
This proof is from John Kelley "General Topology".