My idea here was to use an extremal argument, with two equal length maximum paths on a tree. I said that no matter where the two graphs started, they'd end up having to cross paths since if there were two distinct paths with equal lengths the graph would no longer be connected; the paths still needed to intersect.
I'm just wondering if my explanation is too shallow, especially regarding the usage of the extremal argument. Thank you in advance.
As people said in the comments, the question should additionally say the graph is be connected. We proceed by contradiction. Assume that we have two paths of maximal lengths $m$ which are disjoint. Then all vertices are distinct and thus by connectivity there is a way to get from a vertex A of one path to vertex B of the other, using only edges that are not in the paths.
Now notice that A cuts the first path into two, the longer of which has length at least $\frac{m}{2}$, and similarly for B. Thus it is possible to walk along the longer of the two parts on the A side, along the path connecting A and B, then along the longer of the two parts on the B side to have a path of length strictly more than $m$, a contradiction.
Note: to see why your question is false for unconnected graphs, take the union of two $K_2$s.