Suppose that T1 and T2 are distinct MSTs for an undirected graph G. Let (u,v) be the lightest edge that is in T2 and not in T1. Let (x,y) be any edge that is in T1 and not in T2. What can you say about (u,v) and (x,y)?
Would it be correct to say that the w(u,v) < w(x,y) because adding (u,v) to T1 would create a cycle and thus contradict the MST property?
Take the complete graph $K_4$ with all weights to 1, and two of its spanning trees $T_1$ and $T_2$. Then $w(u,v) = w(x, y)$ for any $uv$ in $T_2$, and any $xy$ in $T_1$, so it cannot be true in general.
In fact if $w(u, v) < w(x, y)$, take $xy$ to be the edge with smallest weight in $T_1$ and not in $T_2$. Then of course $w(x, y) > w(u, v)$ which makes your statement false when you swap $T_1$ for $T_2$.
Another question would be : can you really have $w(x,y) > w(u, v)$, where $xy$ in $T_1$ and $uv$ in $T_2$ are the smallest edges of each tree not in the other ? Well, no. Add $uv$ to $T_1$. This creates a cycle with $uv$ being the lightest edge. Thus removing any edge on the cycle other than $uv$ would give you a smaller spanning tree than $T_1$, a contradiction.
So $w(u, v) \leq w(x, y)$, and if $w(u, v) = w(x, y)$ if and only if $xy$ is the lightest edge in $T_1$ not in $T_2$.