A connected k-regular bipartite graph is 2-connected.

Let $G = (V_{1}\cup V_{2},E)$ be a connected, $k$-regular bipartite graph where $V_{1}$ and $V_{2}$ are the partite vertex sets. As the case $k=1$ is trivial, we may assume that $k \geq 2$ and therefore $|V_{1}\cup V_{2}|\geq 4$.

Assume for contradiction that $G$ is not $2$-connected.

As $G$ is connected but not $2$-connected, there exists a vertex $v$ whose removal disconnects the graph. Without loss of generality we may assume that $v \in V_{1}$. Then $G-v = \uplus_{i\in [1,a]} G_{i}$ where each $G_{i}$ is a connected component and $a \geq 2$.

As $a \geq 2$, there exists some component $G_{b}$ such that $|V_{1}\cap V(G_{b})| \geq |V_{2}\cap V(G_{b})|$. (It shouldn't be too hard to convince yourself of this) For convenience denote $L = V_{1}\cap V(G_{b})$ and $R = V_{2}\cap V(G_{b})$.

As $G_{b}$ is a connected component and $G$ was connected, and $v \in V_{1}$, at least one vertex in $R$ was adjacent to $v$, and therefore has degree less than $k$. However the vertices in $L$ have lost no edges Then we have $$ \sum_{u\in R}deg(u) < k\cdot|R| < k\cdot|L| = \sum_{w \in L}deg(w) $$ However as $G[L \cup R]$ forms a bipartite graph, we know $$ \sum_{u\in R}deg(u) = \sum_{w \in L}deg(w) $$

Thus we have a contradiction, so $G$ must be at least $2$-connected.

If there's a cut-vertex $v$, some neighbor of $v$ induces in $G \setminus \{v\}$ a connected bipartite graph with bipartition $A \cup B$. It looks something like:

We observe:

• By definition, the number of edges from $A$ to $B$ is $k|V(A)| \equiv 0 \pmod k$.
• If $b$ is the number of edges from $B$ to $v$, then $b < k$, otherwise $v$ is not a cut vertex (in this case, deleting $v$ gives the connected graph induced by $A \cup B$).
• By definition, the number of edges from $B$ to $A$ is $k|V(B)|-b \not\equiv 0 \pmod k$, since $k \geq 2$.

This is a contradiction.