This is saying that if $G$ is a connected regular bipartite graph in which each vertex has degree $\geq 2$, and if $e$ is an edge of $G$, then $G \setminus e$ (that is, the graph obtained from $G$ by removing $e$) is still connected. The quickest way to do this might be the following one: A famous corollary of Hall's marriage theorem shows that any regular bipartite graph can be decomposed into perfect matchings. Thus, $G$ can be decomposed into at least $2$ perfect matchings (at least $2$ because each vertex has degree $\geq 2$). Let us decompose $G$ in this way, and pick two of these perfect matchings, one of which contains $e$. But the union of two perfect matchings is always a disjoint union of cycles (another known fact). Thus we conclude that $e$ belongs to a cycle of $G$. This entails that $G \setminus e$ is still connected.

A more direct approach. Let $G(A,B)$ be the corresponding bipartite graph and we remove an edge $e=a_0\,b_0\,, a_0\in A, b_0\in B$. Suppose on the contrary, the resulting graph is not connected and let $G_0(A_0,B_0)$ be its connected component containing $a_0$ i.e. $a_0\in A_0$. Then $b_0\notin B_0$ otherwise there would be a path from $a_0$ to $b_0$ in $G_0$. Since there are no edges connecting vertices of $G_0$ with vertices outside $G_0$, the degrees of all vertices in $G_0(A_0,B_0)$ are $k$ except $a_0$ which degree is $k-1$. Let $|A_0|=n, |B_0|=m$. By counting edges in $G_0$ we get $$nk-1=mk\quad (*)$$But in case $m\ge n$ the RHS of $(*)$ is greater than the LHS, and in case $m<n$ the LHS is greater (because $k\ge 2$). Hence, there is no chance $(*)$ to hold, contradiction.