G is connected, but is not connected if any single edge is removed from G. $\implies$ Any two vertices in G can be connected by a unique simple path. How does how show the following implication to be true:
G is connected, but is not connected if any single edge is removed from G. $\implies$
Any two vertices in G can be connected by a unique simple path.
 A: Suppose that $G$ is connected and there exists a pair of vertices $u,v$ in $G$ such that there exist two simple paths from $u$ to $v$. Can you find an edge to remove so that $G$ is still connected?
A: The way hinted by Michael Biro is probably the simplest, but you could also consider this alternate approach:


*

*Prove by induction that $G$ has exactly $|V|-1$ edges (removing any edge will get you two smaller (connected) graphs matching the hypothesis).

*Any connected graph with exactly $|V| - 1$ edges is a tree (if it has less, then it cannot be connected, if it has more, then it has a cycle).

*In a tree any two vertices are connected by a unique simple path (there are no cycles).


Good luck ;-)
A: Following dtldarek's observation about $G$ being a tree:
Note that is suffices to show that $G$ is a tree.
Keep in mind that a tree is a connected acyclic graph. You already know the graph $G$ is connected. It just remains to show it is acyclic. So you can proceed by contradiction. Suppose $G$ is connected and it has a cycle $C$. You are only left to show that if such cycle $C$ exists, then the hypothesis is no longer true.
