 Solution

We verify the Hall condition for $U$. Let $S\subseteq U$. If $S$ is empty, then $|N(S)|=0=|S|$. Suppose $0<|S|\le n/2$. Choose $u\in S$. Its graph neighbourhood is contained in $N(S)$, so the minimum degree hypothesis givesIf $|S|>n/2$, suppose for a contradiction that some vertex $w\in W\setminus N(S)$ exists. All neighbours of $w$ lie in $U\setminus S$, socontradicting the minimum degree hypothesis. Thus $N(S)=W$ and $|N(S)|=n\ge |S|$.

Every subset $S\subseteq U$ satisfies the Hall condition. By Hall's marriage theorem, there is a matching in a graph saturating $U$. Since $|U|=|W|=n$, its $n$ edges also cover every vertex of $W$. It is therefore a complete matching, meaning a perfect matching. For $n=0$, the empty matching in a graph already has this property.