Let T be a tree with at least two vertices and let $v \in V(T)$. If T-v is a tree, then v is a leaf.
Attempt: Let T be a tree and let v be a leaf of T. Then T-v is a tree.
Let $a,b \in V(T-v)$. We must show that there is an (a,b) path in T-v. We know that since T is connected, there is (a,b)-path P in T. I claim that O does not include the vertex v. Since v is neither the first nor the last vertex on this path, it has two distinct neighbors on the path, contradicting the fact that $d(v)=1.$ Therefore P is an (a,b)-path in t-v and T-v is connected and a tree.
I just proved the converse of my original question. But a little stuck proving original question.