Past exam of the options II course of the University of Oxford / 2008 printed variant / E / E4 / b

Every $K_n$ is closed in the Hausdorff ambient space, because it is compact. Inside the compact space $K_1$, these sets have the finite intersection property: a finite intersection equals the member with largest index and is nonempty. If the total intersection were empty, their open complements would cover $K_1$; a finite subcover would contradict that property. Thus

If no $K_n$ were contained in $U$, the sets $L_n=K_n\setminus U$ would be nonempty, nested and compact, since each is closed in $K_n$. Part (i) would supply a point in ${\bigcap }_n{L}_n$, belonging to every $K_n$ but not to $U$, a contradiction. Thus some $K_{n_0}\subseteq U$, and nesting gives the same containment for every later index.