$C$ set of closed compact sets, $\bigcap_{A \in C} A \subseteq U$. Show there's finite $C' \subseteq C$: $\bigcap_{A \in C'}A \subseteq U$ I'm learning about compact sets in topology and have problems doing the following exercise:
Let $C$ be a collection of closed compact sets in a topological space $X$ and let $U \subseteq X$ be open such that $\bigcap_{A \in C} A \subseteq U$. Show that there exists a finite subcollection $C' \subseteq C$ such that $\bigcap_{A \in C'} A \subseteq U$.
 A: I'm not sure this is helpful, but here is a dull proof using the finite intersection property:
For $A \in C$ and let $A^* = A \setminus U$, note that $A^*$ is compact as well since it is a closed subset of $A$.
Now proceed by contradiction and suppose that for any finite subcollection $I$ that 
$\cap_{A \in I} A $ is not contained in $U$. In particular, this means that $\cap_{A \in I} A^* $ is non empty. Then the finite intersection property states that $\cap_{A \in C} A^* $
is non empty which contradicts $\cap_{A \in I} A \subset U$.
A: Let $V=X\setminus U$, this is closed set. A closed subset of a compact set is compact, hence for each $A\in C$ the set $V\cap A$ (which is closed as an intersection of two closed sets) is compact. Now let's look at the intersection $\cap_{A\in C} (V\cap A)$. It is an intersection of compact sets and it is empty. We will show that this implies there must a finite subcollection $C'\subseteq C$ such that $\cap_{A\in C'} (V\cap A)=\emptyset$. 
First of all we write $C=\{A_i: i\in I\}$. Now let's fix any $j\in I$. For each $i\in I$ let $F_i=X\setminus (V\cap A_i)$, this is an open set. And now note that $\cup_{i\in I} F_i$ is an open cover of $V\cap A_j$. Indeed, if $x\in V\cap A_j$ then there must be some $i\in I$ such that $x\notin V\cap A_i$ (because $\cap_{i\in I} (V\cap A_i)=\emptyset$), and hence $x\in F_i$ for this $i$. Now, $V\cap A_j$ is compact and hence there must be a finite subcover. There is a finite subset $J\subseteq I$ such that $\cup_{i\in J} F_i$ covers $V\cap A_j$. This implies that $(V\cap A_j)\cap (\cap_{i\in J} (V\cap A_i))=\emptyset$. 
So we indeed found a finite $C'\subseteq C$ such that $\cap_{A\in C'} (V\cap A)=\emptyset$. And now it simply implies that $\cap_{A\in C'} A\subseteq U$. 
