Let $G=(V,E)$ be a connected, $d$-regular bigraph. Prove that for any $v\in V$, $G' = G\setminus\{v\}$ is connected.
$G$ is a bigraph, so $V = A\cup B$. $G$ is also $d$-regular, so $|A| = |B|$. I can see why it's true, but I'm struggling proving it. I'll be glad for your help!
In the trivial case $d=1$, we have $|V|=2$ and $|E|=1$, and it is readily solved.
In the case $d=2$, we can show that $G$ is a cycle. In fact, let $V=\{v_1, v_2,\ldots,v_n\}$, where we arrange the vertices in a manner that $v_1, v_2,\ldots,v_k$ is the longest sequence such that $v_1v_2, v_2v_3, \ldots, v_{k-1}v_k\in E$. If $k<n$ then there exists $i < k$ such that $v_iv_k\in E$. It follows from 2-regularity of $G$ that $i=1$, hence the subgraph generated by $v_1,\ldots,v_k$ is a (proper) connected component of $G$, which is contradiction. Hence $k=n$. And $G$ is a cycle. For a cycle it is obvious that $G\backslash\{v\}$ is connected for any $v\in V$.
But it seems difficult to generalize to the cases $d>2$. Besides, the above deduction didn't use the condition "bigraph".