Prove that there is a sequence of k minimum spaning trees between two distinct minimum spanning trees that each one is different in only 1 edge
This is a generalization of the Uniqueness property of minimum spanning trees, and the proof is almost the same:


Note there is at least one edge that belongs to exactly one of $T$ and $T'$. Among such edges, let $e_1$ be the one with least weight. Without loss of generality, assume $e_1 \in T$.
As $T'$ is an MST, $e_1\cup T'$ must contain a cycle $C$ with $e_1$.
As a tree, $T$ contains no cycles, therefore $C$ must have an edge $e_2\notin T$.
Since $e_1$ was chosen as the lowest-weight edge among those belonging to exactly one of $T$ and $T'$, the weight of $e_2$ must be no less than the weight of $e_1$.
As $e_1$ and $e_2$ are part of the cycle $C$, replacing $e_2$ with $e_1$ in $T'$ therefore yields a spanning tree $T''$ with no larger weight, thus also a minimum spanning tree. Since $e_1\in T$ and $e_2\notin T$, $T''$ is more "closer" to $T$ than $T'$.
Repeat the processes above for $T$ and $T''$. Finally you will get a sequence of mimimum spanning trees between $T$ and $T'$.