Let $G$ be a $5$-regular graph with $\kappa(G) = 2$. Prove that $\lambda(G) \leq 4$.

As part of my revision for a graph theory I'm doing through some provided questions and answers, however the answer to the above wasn't provided.

I know that a $5$-regular graph contains vertices all with degree $5$.

So using Whitney's theorem we have: $\kappa(G) = 2 \leq \lambda(G) \leq 4 \leq \Delta(G) = 5$.

I'm just not really sure how to approach proving that Lambda is less than $5$.

If every vertex $v$ has degree $5$ and $\kappa(G) = 2$ then there must be $2$ internally disjoint paths between any $u$ and $v$ in the graph.

because of these two internally disjoint paths, the edges of $u$ must be split between these two paths, as such one path would have $2$ edges and another $3$ edges. This would be the same as $v$, which would at most have $2$ edges from one path and $3$ from the other.

This means that to destroy the connectivity of the graph, you could cut the two edges from $u$ and the two edges to $v$. If one or both of the paths have only $1$ edge, then the number of edges to cut will always be $\leq 4$.

Is this a sufficent proof? Is there anyway of making it simpler/neater?

Thanks