Past exam of the options II course of the University of Oxford / 2013 / E4 / a / ii / 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 provesThis is the nested-closed-set form of the finite intersection property for compact spaces.