Past exam of the topology course of the University of Oxford / 2023 / 1 / a / v

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.