 Solution

The minimum spanning tree cut property, in its safe-edge form, says: if an edge set $A$ is contained in some minimum spanning tree, a cut of a graph respects $A$, and $e$ is a light edge crossing a cut, then $A\cup \{e\}$ is contained in some minimum spanning tree.

Choose an MST $T$ containing $A$. If $e\in T$, nothing changes. Otherwise adding $e$ creates its unique fundamental cycle. The path in $T$ between the endpoints of $e$ must cross the chosen cut; take such an edge $f$. Since the cut respects $A$, $f\notin A$. Since $e$ is light, $w(e)\le w(f)$. The exchange $T-f+e$ is a spanning tree containing $A\cup \{e\}$, with total weight no greater than that of $T$. Minimality of $T$ forces equal optimum weight, proving the claim. Equal edge weights cause no difficulty: “safe” means belonging to some compatible optimum, not to every optimum.