Graph with exactly 2 Minimum Spanning Trees
Lemma: Let $C$ be a cycle of $G$ that contains an unique edge $e$ of maximum weight. Edge $e$ does not belong to any MST of $G$.

Proof:
Suppose that a MST $T^* = (V, E^*)$ of $G$ contains $e = (u,v)$.
Root $T^*$ in $u$ and let $f$ be any edge of $C \setminus E^*$ that has exactly one endpoint in the subtree of $T^*$ rooted in $v$ (this edge always exists since $C \setminus \{ e \}$ is a path from $v$ to $u$ that avoids $e$). The edge $f$ closes a fundamental cycle containing $e$ and is such that $w(f) &lt; (e)$.
Then $(V, (E^* \setminus \{e\}) \cup \{ f \}$) is a spanning tree of $G$ that weighs less than $T^*$. This is a contradiction. $\square$

Let $T_1 = (V, E_1)$ and $T_2 = (V, E_2)$ be two distinct MSTs of $G$.
Let $C$ be a cycle in $(V, E_1 \cup E_2$).
Let $M = \arg\max_{e \in C} w(e)$.

If $|M|&gt;1$ we are done. Suppose then that $M = \{ e \}$.
By the above lemma, $e$ is the unique heaviest edge of $C$ and hence it cannot belong to any MST of $G$. This is a contradiction since $e$ must belong to at least one of $E_1$ and $E_2$.