Theorem. If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty, then $\cap K_\alpha$ is nonempty.
Proof. Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$ [this denotes the complement of $K_\alpha $]. Assume that no point of $K_1$ belongs to every $K_\alpha$. Then the sets $G_\alpha$ form an open cover of $K_1$ [this took me a bit but that's by the last assumption]; and since $K_1$ is compact, there are finitely many indices $\alpha_1,\dots,\alpha_n$ such that $K_1\subset \bigcup_{i=1}^nG_{\alpha_i}$ [so far so good]. But this means that $K_1\cap K_{\alpha_1}\cap\dots K_{\alpha_n} $is empty, in contradiction to our hypothesis.
I just don't see how the very last part is implied. Can someone help me see it?