I bumped into this question in an old exercice sheet :
Let $ G = (V,E) $ be a connected graph with minimum degree $ \delta(G) = k \geq 2$. Prove that there exist a path $P = x_1x_2...x_k$ such that $G \setminus P $ is also connected $G \setminus P$ is the same as $G[V\setminus (x_1,x_2,...x_k)]$.
I wanted to prove it by induction but even for the cas $k=2$ I didn't know where to begin.
I thought maybe using the fact that in a graph with minimum degree k there always exist a cycle of length $k+1$ but then I have no idea how to show that it's possible to remove a part of the cycle.
Thank you for your help