Suppose $T$ is a tree and $u,v$ are vertices s.t there are 2 distinct paths between them.
i.e. $u,u_1,u_2,...,u_n, v$ and $u,u'_1,u'_2,...,u'_m, v$.
If $\forall u_i$, $u_i\notin \{u'_1,u'_2,...,u'_m\}$ this implies $u,u_1,..,v,u'_m,..,u$ is a cycle in T. Contradiction
If not let $I$ be the least index s.t $u_i\in \{u'_1,u'_2,...,u'_m\}$.
Therefore, $u,u_1,u_2,...,u_I,u'_{I-1},u'_{I-2},...,u$ is now a cycle in T. Contradiction.
Hence, there is a unique path between any 2 vertices.
For the other direction suppose for contradiction there is a cycle in T i.e. $u,u_1,u_2,...,u_n,u$. Then, $u,u_1$ and $u,u_n,u_{n-1},..,u_1$ are 2 distinct paths in T.
Are the above proofs correct?
Nearly. Remember that a tree is both acyclic and connected. You showed that a graph with a cycle contains two vertices that are not connected by a unique path, but you also need to show that a disconnected graph also contains two vertices that are not connected by a unique path.