Applying the contra positive of the finite intersection property I'm reading a proof which has the following setting. 
I have a family $D$ of compact sets with empty intersection.  The next line takes a finite subset of $D$ with empty intersection.
This is clearly possible if families of compact sets enjoy the finite intersection property.  Upon trying to verify this I realize it might be much harder than it sounds.  Is this the case or am I barking up the wrong tree?
 A: It’s true in any Hausdorff space. (Hausdorffness ensures that compact sets are closed.)
Let $\mathscr{K}$ be a family of compact sets in a space $X$, and suppose that that $\bigcap\mathscr{K}=\varnothing$. Fix $K_0\in\mathscr{K}$, and let $\mathscr{K}_0=\mathscr{K}\setminus\{K_0\}$. For each $K\in\mathscr{K}_0$ let $U_K=X\setminus K$, and let $\mathscr{U}=\{U_K:K\in\mathscr{K}_0\}$. Then $\mathscr{U}$ is an open cover of $K_0$, and $K_0$ is compact, so $\mathscr{U}$ has a finite subcover. Let $\mathscr{F}$ be a finite subset of $\mathscr{K}_0$ such that $\{U_K:K\in\mathscr{F}\}$ covers $K_0$; then $\mathscr{F}\cup\{K_0\}$ is a finite subset of $\mathscr{K}$ with empty intersection.
A: Brian's answer is quite nice and complete, just slightly different way of looking at it: suppose we know that compactness is equivalent to "every family of closed sets with the FIP has non-empty intersection". 
In your case you have a family $\mathcal{K}$ of compact sets with empty intersection. Then taking one compact member $K_0 \in \mathcal{K}$ the family $\mathcal{K}' = \{ K \cap K_0: K \in \mathcal{K} \}$, the so-called trace of $\mathcal{K}$ on $K_0$, also has empty intersection (it's the same intersection as the original family had..) and now is a family of closed (here we use Hausdorff) sets in a compact space ($K_0$) that has empty intersection. 
By the contrapositive of the characterisation, $\mathcal{K}'$ does not have FIP, so some finite subfamily has empty intersection, and gives us a finite subfamily of $\mathcal{K}$ with that property as well.    
