Let $G'=(V',E')$ be a counterexample with the minimum number of vertices. Note that $|V'| \geq 3$. By choice of $G'$,
(1) For every vertex $v \in V(G')$, we have that $\delta(G'-v) = \delta (G')-1$.
Let $D$ be the set of minimum degree vertices of $G'$. By (1), each vertex in $D$ has degree $|V'|-1$ in $G'$. This implies that $D=V'$, and that $G'$ is a clique. Since $G'$ is bipartite, we have that $|V'| \leq 2$, a contradiction.