Prove that every two longest paths have at least one vertex in common

EDIT Updated slightly to make explicit that $P’$ may only be of length one. Assume for contradiction that $P_{1} = langle v_{0},ldots,v_{k}rangle$ and $P_{2} = langle u_{0},ldots,u_{k}rangle$ are two paths in $G$ of length $k$ with no shared vertices. As $G$ is connected, there is a path $P’$ connecting $v_{i}$ to $u_{j}$ for some $i,j in [1,k]$ such that $P’$ shares no vertices with $P_{1} cup P_{2}$ other than $v_{i}$ and $u_{j}$. Say $P’ = langle v_{i},x_{0},ldots,x_{b},u_{j}rangle$ (note that there may be no $x_{i}$ vertices, i.e., $b$ may be $0$ – the notation is a bit deficient though). Without loss of generality we may assume that $i,j geq lceilfrac{k}{2}rceil$ (we can always reverse the numbering). Then we can construct a new path $P^{*} = langle v_{0},ldots,v_{i},x_{1},ldots,x_{b},u_{j},ldots,u_{1}rangle$ (by going along $P_{1}$ to $v_{i}$, then across the bridge formed by $P’$ to $u_{j}$, then along $P_{2}$ to $u_{0}$). Obviously $P^{*}$ has length at least $k+1$, but this contradicts the assumption that $G$ has no paths of length greater than $k$. So then any two paths of length $k$ must intersect at at least one vertex and your observation that it must be in the middle (if there’s only one) follows as you reasoned.