Necessity. If $M$ saturates $X$, then for any $S \subseteq X$, the matching maps $S$ injectively into $N(S)$, so $|N(S)| \geq |S|$.

Sufficiency (by induction on $|X|$). Base case $|X| = 1$: Hall's condition gives $|N(\{x\})| \geq 1$, so $x$ has a neighbor, and the single edge is the desired matching.

For $|X| > 1$, distinguish two cases.

Case 1: $|N(S)| \geq |S| + 1$ for all $\emptyset \neq S \subsetneq X$ (strict inequality for proper subsets). Pick any edge $xy$ with $x \in X, y \in Y$. In $G' = G - \{x, y\}$ (remove $x$ and $y$), for any $S' \subseteq X \setminus \{x\}$: $|N_{G'}(S')| \geq |N_G(S')| - 1 \geq (|S'| + 1) - 1 = |S'|$. By induction, $G'$ has a matching $M'$ saturating $X \setminus \{x\}$, and $M = M' \cup \{xy\}$ saturates $X$.

Case 2: There exists $\emptyset \neq T \subsetneq X$ with $|N(T)| = |T|$. Consider the subgraph $G_1$ induced by $T \cup N(T)$. Hall's condition holds for $G_1$ (since $N_{G_1}(S) = N_G(S)$ for $S \subseteq T$, as neighbors of $T$-vertices in $Y$ lie in $N(T)$). By induction (on $|T| < |X|$), $G_1$ has a matching $M_1$ saturating $T$.

Now consider $G_2 = G - (T \cup N(T))$, a bipartite graph on $(X \setminus T) \cup (Y \setminus N(T))$. For $S \subseteq X \setminus T$: $|N_{G_2}(S)| = |N_G(S \cup T)| - |N(T)| \geq |S \cup T| - |T| = |S|$ where we used Hall's condition for $S \cup T$ in $G$ and $|N(T)| = |T|$.

By induction, $G_2$ has a matching $M_2$ saturating $X \setminus T$. Then $M = M_1 \cup M_2$ saturates all of $X$. $\square$