Question: I have the following problem: 

Let $X$ be a Hausdorff Space, and $J \neq \varnothing$, such that $\{K\_{j}:j \in J\}$ is a family of compact sets, whereby any finite subset $J\_{0} \subseteq J$ shows that $\cap\_{j \in J\_{0}}K\_{j} \neq \varnothing$. Prove that $\cap\_{j \in J}K\_{j} \neq \varnothing$.

My idea: Let's assume that $\cap\_{j \in J}K\_{j} = \varnothing$. We now look at $K\_{j}^c:=X-K\_{j}$. It is clear that $K\_{j}^c$ is open as the $K\_{j}$ is a compact subset in a Hausdorff Space. We can now say that $X=\cup\_{j \in J\_{0}}K\_{j}^{c}$. 

**This is where my question comes in: There was no mention made of the compactness of $X$. Is there any way of proving the above without compactness of $X$?** 

Using compactness, I would say: $\exists N \in \mathbb N: \cup\_{j=1}^{N}K\_{j}^{c}=X$, and that therefore $\cap\_{j=1}^{N}K\_{j}^{c}=\varnothing$ which is a contradiction. q.e.d

Answer: 
Since there are four options for each choice, and each choice is independent of the others, and there are two choices, there are $4^2=16$ total options.
You take one and you have got 3 options left. So three possibilities if you pick a first, another three if you you pick b first .... . So in total you have got 12 possibilities.