Solution
= Solution
If the intersection were empty, the open complements $X\setminus F_i$ would cover the <compact> space. A finite subcover would give an empty finite intersection of the corresponding $F_i$. But the sets are nested, so any finite intersection is the set with largest index, which is nonempty. This contradiction proves
$$
\boxed{\bigcap_iF_i\ne\varnothing.}
$$
This is the nested-closed-set form of the <finite intersection property> for <compact> spaces.