Let G be k-reguler bipartite graph of degree at least 2. show that K(G) not equal 1?
The statement is false. Here is a counterexample. Let $E$ be the even numbers and $O$ be the odd numbers, and Let $N=\{(n,n+1)|n\in\mathbb{N}\}$. Then $G=(E\cup O, N)$ is a $2$-regular, bipartite graphs, every vertex of which is a cut-vertex.

I won't do your homework for you until you can show where you get stuck, but I'll give you a hint: the conjecture is true for finite graphs, and the proof hinges on this fact. You say, well the resulting components have some number of vertices, and these vertices of one component are distributed between the partitions, so I can count the number of edge ends on either side of the partitions and then you double check if it comes out to the same. It also hinges on the fact that the two partitions are equally big: for a $k$-regular graph $G=(U\cup V, E)$, $|U|=|V|$, a fact which you will have to prove first. The proof for that hinges on the fact that the graph is regular.