If G is a graph such that any two vertices of G are connected by exactly one path, then G is a tree.

I am thinking about something along these lines:

Assume that $$G$$ is connected and let $$u$$ and $$v$$ be two vertices that are disjoint. In order for $$u$$ and $$v$$ to be connected there exists another vertex $$w$$ such that $$uw$$ and $$wv$$ are connected. Now since $$u$$ and $$v$$ are both connected with $$w$$ then $$G$$ is a tree.

Does that work OR is it more complicated than that?

• Did you mean "Assume $G$ is connected?" – angryavian Apr 11 at 0:32
• @angryavian Yes correct! sorry about that – Hidaw Apr 11 at 0:35

You would need more than just there be a connected graph on $$uw$$ and a connected graph on $$wv.$$ Here is a way I would go about it.
Since every vertex of $$G$$ is linked by a path, $$G$$ is connected. If $$G$$ had a cycle, and $$u,w$$ were on that cycle, then they would be connected by two distinct paths. This cannot happen, so $$G$$ has no cycles. The only connected acyclic graphs are trees, hence $$G$$ is a tree.