Rudin Analysis, Theorem 2.36: Is there a generalization? In Rudin's Mathematical Analysis, he states theorem a theorem for compact sets. 

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 $\bigcap K_\alpha$ is nonempty.

The proof presented in the text uses the definition of compacts sets to contradict the negation of the hypothesis.
My question is: do the sets need to be compact?
It seems that you can present this theorem with a collection of general sets.  The proof may be gone about by assuming that a set $K_1$ has elements that are not in every set.  Then create a contradiction using the the hypothesis: that every finite subcollection of sets is non empty.
Could you build a proof around that?  If so how?
If it only works for compact sets, I would be interested in finding an example of non-compact sets where the theorem breaks down.
 A: Consider the metric space $\mathbb R$ with the usual metric.  The open intervals $(0,1/n)$ for $n \in \mathbb N$ comprise a collection of subsets of $\mathbb R$.  The intersection of every finite subcollection is nonempty [choose the greatest $n$ for which $(0,1/n)$ is in the finite subcollection and then $1/(n+1)$ is in the intersection of the finite subcollection] but the intersection of the entire collection is empty [if it contained $x$ we would have $0<x<1/n$ for all $n$, violating the Archimedean property].  
The unbounded intervals $[n, \infty)$ for $n \in \mathbb N$ comprise another collection of subsets of $\mathbb R$ where the intersection of every finite subcollection is nonempty [choose the greatest $n$ for which $[n, \infty)$ is in the subcollection, and the intersection is $[n, \infty)$] but the intersection of the entire collection is empty [if it contained $m$ we would have $m>n$ for all $n$, which is absurd.
In summary, the conclusion of the theorem does not hold for collections of subsets of $\mathbb R$ that are not compact (closed and bounded).
A: It is enough that all subsets are closed and at least one of them is compact, the space $X$ itself can be any topological space. Under those conditions such a family that has the FIP (finite intersection property) has non-empty intersection. 
