Let $\{C_j:j \in J\}$ be a family of closed compact subsets of a topological space $(X,\tau)$. Prove that $\bigcap_{j \in J}C_j$ is compact.
I realized this is not a metric space, so compactness in general topology does not imply closed or boundedness. But if we use the subcover definition of compactness, it should always by possible find a finite open subcover right since each $C_j$ is compact and each has a finite subcover, the intersection of finite open subcover is still open and is still a subcover for the intersection of all the compact sets.
One concern I have: even though each $C_j$ has a finite subcover, if there are an uncountable number of $C_j$, then we are taking an infinite intersection of open subcovers which is not open.
Best Answer
Perhaps worth adding this example for why "closed" is a necessary assumption here.
Let $X=\mathbb Z\cup\{-\infty,\infty\}$ have the topology $\{U\subseteq X:U\cap\{-\infty,\infty\}\not=\emptyset\Rightarrow \mathbb Z\subseteq U\}$. Then $\mathbb Z\cup\{\infty\}$ and $\mathbb Z\cup\{-\infty\}$ are compact, not closed, and their intersection is the infinite discrete (and thus not compact) subspace $\mathbb Z$.
(Such a space is $T_0$, but not $T_1$. However, it can be modified to be $T_1$: use $\{U\subseteq X:U\cap\{-\infty,\infty\}\not=\emptyset\Rightarrow \mathbb Z\setminus U \text{ finite}\}$ instead.)
Since this was marked as the accepted answer, here's a writeup answering the question.
Let $0\in J$ and consider the compact subset $C_0\subseteq X$. Then for $j\in J$, note that $C_0\cap C_j$ is closed in the subspace topology for $C_0$. Then $\bigcap_{j\in J} C_j=\bigcap_{j\in J}(C_0\cap C_j)$ is the intersection of closed subsets of the space $C_0$, and is therefore closed. Since a closed subset of a compact space is compact, we have $\bigcap_{j\in J} C_j$ compact.
(Note that I do assume $J$ is non-empty by letting $0\in J$ - see user14111's comment.)