 Solution

A matching from $X$ to $Y$ is a set of pairwise vertex-disjoint edges that covers every vertex of $X$. Equivalently, it chooses for each $x\in X$ a distinct neighbour in $Y$.

Hall marriage theorem says that such a matching exists if and only ifNecessity is immediate: the distinct partners of the vertices in $S$ all belong to $N(S)$.

For sufficiency, induct on $|X|$. The claim is clear when $|X|=1$. First suppose every nonempty proper $S\subset X$ satisfies the strict inequality $|N(S)|\ge |S|+1$. Choose an edge $xy$ and delete $x$ and $y$. For $S\subseteq X\setminus \{x\}$, deletion removes at most one neighbour, soInduction gives a matching of $X\setminus \{x\}$, and adding $xy$ completes it.

Otherwise there is a nonempty proper $S\subset X$ with $|N(S)|=|S|$. Hall's condition holds in the bipartite graph induced by $S\cup N(S)$, so induction matches $S$ onto $N(S)$. For $A\subseteq X\setminus S$,and hence $A$ has at least $|A|$ neighbours outside $N(S)$. Induction therefore matches $X\setminus S$ into $Y\setminus N(S)$. The two matchings are disjoint and together cover $X$.