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 $\bigcap_nL_n$, belonging to every $K_n$ but not to $U$, a contradiction. Thus some $K_{n_0}\subseteq U$, and nesting gives \b[the same containment for every later index].

Solution

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

 Solution (source code)