# Counterexample finite intersection property

Let $$(X,\mathcal T)$$ be a (not necessarily Hausdorff) topological space. It is well-known that $$X$$ is compact (in the sense of every open cover has a finite subcover) if and only if for every family of closed subsets $$(C_i)_{i\in I}$$ of $$X$$ satisfying the finite intersection property, $$\bigcap _{i\in I}C_i$$ is nonempty. Now does the following assertion hold:

If $$(X,\mathcal T)$$ is compact and $$(C_i)_{i\in I}$$ is a family of $$\textbf{compact}$$ subsets satisfying the finite intersection property, then $$\bigcap_{i\in I}C_i$$ is compact.

Since compact subsets in non-Hausdorff spaces need not be closed, I am not sure if this holds. Is there a counterexample?

This is not true in general. For instance, let $$X=[0,1]$$ with the topology that a set is open iff it is downward closed (i.e., $$x\in U$$ and $$y\leq x$$ implies $$y\in U$$). Then $$X$$ is compact since any open set containing $$1$$ is the whole space. Now for $$i\in (1/2,1]$$, let $$C_i=[0,1/2)\cup(1/2,i]$$. Each $$C_i$$ is similarly compact, and they have the finite intersection property. But their intersection $$[0,1/2)$$ is not compact, since it is covered by the open sets $$[0,r)$$ for $$r<1/2$$.
Take the set $$X=\mathbb{N}\cup\{a_1,a_2\}$$ where $$a_1,a_2$$ are some points outside of $$\mathbb{N}$$. We define the following topology: a set is open if it is $$\mathbb{N}\cup\{a_1\},\mathbb{N}\cup\{a_2\}$$, $$\mathbb{N}\cup\{a_1,a_2\}$$ or any subset of $$\mathbb{N}$$. It is easy to see that this is a compact topological space and $$\mathbb{N}\cup\{a_1\},\mathbb{N}\cup\{a_2\}$$ are compact subsets with non empty intersection. However, their intersection is $$\mathbb{N}$$ which is not compact.