I want to prove the folloeing:
Let $G$ be a cubic ($3$-regular) $3$-connected graph and $x,y\in V(G)$. Then there is a $x-y$ path $P$ in $G$ such that $G-V(P)$ is connected.
I tried to prove it as follows:
Since $G$ is $3$-connected, there are three disjoint $x-y$ paths, which I want to take as candidates for $P$. Removing one of them and calling it $P$, there are still two disjoint paths $P_1,P_2$ between the two remaining neighbors of $x$ and $y$, so if I prove that for every $a\in V(G)-V(P)$ there is a path to each of $P_1$ and $P_2$, disjoint with $P_3$, I'll be done.
But I don't know if that's even true.
Is there any hope for this try, or a better way to prove it?