In every bipartite graph G, there exists a vertex v such that v is matched in every maximum matching WITHOUT Gallai? An essential vertex is a vertex that is included in all maximum matchings of a graph. And in any connected bipartite graph such a vertex exists. The proof is given in this answer. But this proof is through Gallai's theorem. Is there any other way of proving the same? For instance, using Konig's theorem or Hall's theorem?
Any help would be appreciated.  
 A: Let $G$ be a connected bipartite graph with bipartition $\{A,B\}$. (We assume that $G$ has more than one vertex, and therefore $A$ and $B$ are nonempty, otherwise the claim is false.) If $G$ contains a matching of $A$, then every vertex of $A$ is essential. Otherwise, by Hall's theorem, there exists $S\subseteq A$ such that $|N(S)|<|S|$. Choose a smallest such $S$. We claim that every vertex of $N(S)$ is essential, thus completing the proof because $G$ connected with more than $1$ vertex and $|S|>0\implies|N(S)|>0$.
Let $v\in N(S)$ and suppose a matching $M$ in $G$ does not contain $v$. We find an augmenting path starting at $v$, showing that $M$ is not maximum. Let $H$ be the subgraph of $G$ induced by $S\cup N(S)$. Let $X$ be the set of vertices of $N(S)$ that can be reached from $v$ by an alternating path contained in $H$, and $Y$ the set of penultimate vertices on these paths. If $Y=\varnothing$, then $v$ has an unmatched neighbor in $X$, so their shared edge is an augmenting path.
Suppose instead $Y\neq\varnothing$. $M$ provides a bijection of $X\setminus\{v\}$ and $Y$, so $|Y|=|X|-1\le N(S)-1<|S|-1\implies Y\subsetneq S$. Then there are adjacent vertices $x\in X$ and $u\in S\setminus Y$—otherwise,
$$N(S\setminus Y)\subseteq N(S)\setminus X\implies|N(S\setminus Y)|\le|N(S)|-|X|<|S|-|Y|=|S\setminus Y|\,,$$
so $S\setminus Y$ would contradict the minimality of $S$.
Let $P$ be an alternating path from $v$ to $x$. $u$ is not matched—otherwise, we could extend $P$ by the unmatched edge $xu$ and the matched edge from $u$, creating an alternating path starting from $v$ with penultimate vertex $u$, but $u\notin Y$. Because $u$ is not matched, we can extend $P$ by the unmatched edge $xu$ to create an augmenting path, completing the proof.
