 Solution

Because $T$ itself is a tree, every finite induced connected exhaustion has the unique spanning tree consisting of all its edges. Thus its free spanning forest is deterministically $T$.

By the stated transience criterion, choose an edge $e=uv$ whose two complementary subtrees are transient. Run Wilson algorithm rooted at infinity first from $u$. With positive probability its loop-erased walk remains forever in the $u$-side. Starting next from $v$, there is likewise positive conditional probability that its walk remains forever in the $v$-side. On this event the two rays never use $e$, so $e$ is absent from the wired uniform spanning forest. The wired law is therefore not the deterministic free law.