Is every edge of a graph included in some spanning tree?

Let's say we have a graph $G$. We pick one edge from it (any edge). Will there always be such a spanning tree that contains that very edge?

I think the answer is yes, because no matter what we do we can always create such a spanning tree so that the very edge we picked is included. Of course, a more formal proof would be needed?

Note that the initial graph $G$ needs to be connected or it has no spanning trees at all. (Though the same argument applied to each component would show that any graph has a spanning forest containing any chosen edge.)

Let $G=(V,E)$ be a connected graph, let $T$ be any spanning subtree and let $e$ be any edgein $E$. We claim that there is a spanning tree that includes $e$. If $e\in T$, we are done. Otherwise, $T+e$ contains a cycle. That cycle necessarily contains at least one edge $e'\neq e$ (actually, it contains at least two). $T+e-e'$ is a spanning tree that contains $e$.

You should prove to yourself that $T+e-e'$ really is a spanning tree.

StackExchange Computer Science Q#87304, answer score: 5