= Solution
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> $A=B\sqcup C$. 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> satisfies
$$
|d(x,F)-d(z,F)|\leq d(x,z).
$$
Indeed, the <triangle inequality> gives $d(x,F)\leq d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two <distance to a set> functions below are <continuous>. Set
$$
U=\{x:d(x,B)<d(x,C)\},\qquad V=\{x:d(x,C)<d(x,B)\}.
$$
These 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>.
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