# Every closed subspace of a compact space is compact using finite intersection property

My book states that the fact that "every closed subspace of a compact space is compact" is a consequence of the following theorem:

A Hausdorff space is compact if and only if every family of closed subsets of $X$ which has the finite intersection property has non-empty intersection.

For completeness: A family $\mathcal{F} = {F_s}_{s\in S}$ has the finite intersection property if $\mathcal{F} \ne \emptyset$ and the intersection of any finite number of the $F_s$ is non-empty.

I'm having trouble seeing how to use this theorem. Should we apply it to the whole space (which is indeed Hausdorff since being Hausdorff is a requirement for compactness in my book), or whether to apply it to the closed set in question.

Let $A$ be a closed subspace of a compact space $X$. To conclude that $A$ is compact, we show that if a collection $\mathcal{F}$ of closed subsets of $A$ has the finite intersection property, then $\cap\mathcal{F}$ is nonempty. Since $A$ is closed, a closed subset in $A$ is also closed in $X$. Thus $\mathcal{F}$ is a collection of closed subsets of $X$ with the finite intersection property. By the compactness of $X$ we conclude $\cap\mathcal{F}$ is nonempty.