Prove that a $k$-regular bipartite graph with $k \geq 2$ has no cut-edge
Ok, here is a simple proof I came up by myself. I was apparently going the wrong way in trying to prove the exercise.

Proof. Suppose edge $e$ is a cut-edge. Study graph $G-e$. For any $v \in G-e$ other then $e$ endpoints $x, y$, the vertex degree $d(v) = k$, and $d(x)=d(y)=k-1$. 

Now, since graph $G$ is bipartite, graph $G-e$ remains bipartite. Also, because $e$ is a cut-edge, $G-e$ is composed of 2 components $H_1, H_2$, which are also both bipartite, and each contains exactly one of $x, y$. Assume WLG that $H_1$ has vertex $x$, that $H_1$ has vertex bipartition $X_1, Y_1$ and that $x \in X_1$. 

Clearly, in subgraph $H_1$, we have 

$S_1 = \sum_{v\in X_1 }d(v) = k(|X_1|-1) + k-1$ and $S_2 = \sum_{v\in Y_1 }d(v) = k(|Y_1|)$, and $S_1$ can't be equal $S_2$ unless $k=1$. However, it must be the case that $S_1 = S_2$ in a bipartite graph. Thus, our initial assumption that $e$ is a cut-edge was wrong. Therefore, $G$ has no cut-edge. $\Box$