= 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, giving
$$
\boxed{T\setminus v\text{ is a tree}.}
$$
This includes the two-<vertex> case, where deletion leaves the one-<vertex> <tree>.
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