Is every edge of a graph included in some spanning tree?
Note that the initial graph&nbsp;$G$ 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&nbsp;$E$. We claim that there is a spanning tree that includes&nbsp;$e$. 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&nbsp;$e$.

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