 Solution

Deleting a vertex and its incident edge cannot create a graph cycle, so $T\setminus v$ is acyclic. It remains to establish connectedness and nonemptiness.

Since the leaf of a tree $v$ has one neighbour, $T$ has at least two vertices, so $T\setminus v$ is nonempty. For any two remaining vertices $a,b$, take their unique simple graph path in $T$. It cannot pass through $v$: an internal vertex on such a graph path has two incident edges, whereas $v$ has vertex degree one. Neither endpoint is $v$, so the entire graph path remains after deletion. The remaining graph is connected and acyclic, givingThis includes the two-vertex case, where deletion leaves the one-vertex tree.