Theorem from Baby Rudin (Thm. 2.36)

Theorem. If $\displaystyle \{K_{\alpha}\}$ is a collection of compact subsets of a metric space $\displaystyle X$ such that the intersection of every finite subcollection of $\displaystyle \{K_{\alpha}\}$ is nonempty, then

$\displaystyle \bigcup_{\alpha}K_{\alpha}$ is nonempty.

I try and prove theorems my self before reading them (seldom looks pretty), and the one I came up with for this theorem is the following.

Theorem 2.34 says that compact subsets of metric spaces are closed. From this I know that every set in $\displaystyle \{K_{\alpha}\}$ is closed.

Theorem 2.24(b) says that for any collection $\displaystyle \{K_{\alpha}\}$ of closed sets, $\displaystyle \bigcap_{\alpha}K_{\alpha}$ is closed.

Now comes my question. Since the empty set is both closed and open, can I just say that $\displaystyle \bigcap_{\alpha}K_{\alpha}$ is nonempty since it is closed? I am pretty sure the answer is no, but it would be nice if someone could explain this a bit.

Thanks.