Ok, let me see if I can get a handle on this. One knows that compact subspaces of metric spaces are closed-true. Next, one knows that the arbitrary intersection of closed sets is closed-true. So your argument is that by these two theorems one would have that $\displaystyle \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha$ is a closed subspace of $\displaystyle X$ and thus can't be empty since $\displaystyle \varnothing$ is both open and closed? Unfortunately this is not the case :|. It's often true that a subspace of a metric space is both open and closed (clopen) for it not to be in general the space would have to be

connected. It's often impossible (in full generality) to complete arguments such as this by looking at openess/closedness (although it works sometimes, like proving things like the fundamental lemma of path connectedness in alg. top.).

For the argument I'll try to give you two well placed hints:

Hint 1: If $\displaystyle \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha=\varnothing$ then $\displaystyle \displaystyle \bigcup_{\alpha\in\mathcal{A}}\left(X-K_\alpha\right)=X$.

Hint 2: If for every finite $\displaystyle \mathcal{B}\subseteq\mathcal{A}$ one has that $\displaystyle \displaystyle \bigcap_{\beta\in\mathcal{B}}K_\beta\ne\varnothing$ then for every finite $\displaystyle \mathcal{B}\subseteq\mathcal{A}$ one has $\displaystyle \displaystyle \bigcup_{\beta\in\mathcal{B}}\left(X-K_\beta\right)\subsetneq X$.

So, since the $\displaystyle K_\alpha$ are closed we know that the $\displaystyle X-K_\alpha$ are open and so the above shows we've found a __member__ with no __brain__.