If $(X,d)$ is any metric space and $\{A_\alpha\}_{\alpha\in I}$ is a collection of nonempty compact subsets of $X$ such that the intersection of any finite subcollection of sets is non empty does that imply that $\bigcap A_\alpha $ is non empty?
I tried to prove this by assuming that it is empty and using the complements of $ A_\alpha $ as an open covering but wasn't sure how to get a contradiction if the statement is true in the first place.
Any help would be appreciated.
Suppose that $\bigcap_{\alpha\in I} A_{\alpha}$ is empty. Then, by De Morgan's law, $X=\bigcup_{\alpha\in I} A_{\alpha}^c$. Pick any $\alpha_0\in I$ (tacitly assuming that the index set $I$ is not empty). It follows that $A_{\alpha_0}\subseteq \bigcup_{\alpha\in I} A_{\alpha}^c=X$. Since $A_{\alpha_0}$ is compact and $\{A_{\alpha}^c\}_{\alpha\in I}$ is an open cover of it, there exist some $n\in\mathbb N$ and $\alpha_1,\ldots,\alpha_n\in I$ such that $A_{\alpha_0}\subseteq \bigcup_{j=1}^n A_{\alpha_j}^c$. But this implies that $\bigcap_{j=0}^n A_{\alpha_j}=\varnothing$, a contradiction to the assumption that finite intersections are not empty.