If $P$ and $Q$ are paths in a connected graph that have no vertices in common, then there exist vertices $u$ and $v$ and a path $P′$ such that $u$ is on $P$, $v$ is on $Q$, $P′$ is a $u–v$ path, and no internal vertex of $P′$ is on either of the paths $P$ or $Q$.
Let $P = (u_1, \ldots u_n)$ and $Q = (v_1, \ldots, v_n)$. Then, let $P' = (u_j, u_{n + 1}, v_{n + 1}, v_j)$ where $1 \le j \le n$.
Would that prove the claim?
Hint: consider $T=(w_1,\dots,w_m)$ a path of minimal length so that $w_1$ lies in $P$ and $w_m$ lies in $Q$. Using minimality, can you show that this path satisfies the desired property?