I am not sure about my proof that goes:
Use induction on the number of vertex of caterpillar graph, C.
Base case, C with n=1 holds since it is a adjacent to no vertex. So the claim holds.
Inductive step. Suppose C with n=1, ... ,k holds. Suppose a C with n=k+1.
C is either a path or a tree. So if it is a tree take a leaf and consider a subgraph with k vertices. It is connected and is a C. So the claim hold by the inductive step. Suppose the the leaf taken away was connected to a node that is not a leaf in the subgraph. Then the claim holds for the original graph (since for every vertex in the subgraph the claim holds). Suppose it was connected to a leaf in the subgraph. Then the leaf with the connection is adjacent to at most one non-leaf vertex. So the claim holds.
If C is a path by definition of path every vertex is adjacent to at most two vertex hence the claim holds.
Hence by induction the claim holds.
Can someone verify? Comments?
Overall, it looks good to me.
I am going to be a little nit-picky here though. You are using the inductive hypothesis, not the inductive step. The inductive step is when you assume the inductive hypothesis and use it prove true for the next case.
Also, you could try for a more contradiction argument. A caterpillar graph is a graph such that removal of the endpoints leaves you with a path. And so we prove by contradiction. Suppose there exists a vertex $v$ in $C$ such that $v$ is adjacent to at least $3$ non-leaf nodes. Now remove the endpoints of $C$. As $v$ still has degree at least $3$, we don't have a path, as every vertex in a path has degree at most two. Hence, we have a contradiction.