Counterexample finite intersection property

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.