A few of us have been looking to prove this but we've been struggling. We've found some proofs online that prove in the wrong direction (if T has unique path then T is transitive).
If you could point us in the right direction or just give us something to work with that would be great just we're completely stumped.
Thanks everyone.
By the comments, you're satisfied that $T$ has a Hamiltonian path; you just don't know that it's unique.
Suppose $T$ has multiple distinct Hamiltonian paths; label the vertices so there are two paths $H_1=p_1\dots p_n$, $H_2=p_{\pi(1)}\dots p_{\pi(n)}$ where $\pi$ is a non-identity permutation.
Since $T$ is transitive, there is an arc from $p_i$ to $p_j$ whenever $j > i$. Since $\pi$ is a non-identity permutation, there must be some $i, j$ such that $j > i$ but $\pi(i) > \pi(j)$. Can you obtain a contradiction?