Exercise 1.4 proof

If there are two different MST $T_a, T_b$, there must exists at least 4 edges with different weight $e_{a_1}, e_{a_2}, e_{b_1}, e_{b_2}$, where

$$e_{a_1}, e_{a_2} \in T_a, e_{a_1}, e_{a_2} \notin T_b, e_{b_1}, e_{b_2} \in T_b, e_{b_1}, e_{b_2} \in T_a$$
and $$e_{a_1} < e_{b_1} < e_{b_2} < e_{a_2}$$
s.t. $$|e_{a_1}|+|e_{a_2}|=|e_{b_1}|+|e_{b_2}|$$
Obviously, these four edges connect with each other, i.e. these four edges forms a cut.
However, from the Cut Lemma, we could simply choose $e_{a_1}$ and $e_{b_1}$, which will make the total weight smaller than $T_a$ and $T_b$, i.e. $T_a$ and $T_b$ is not MST, where we find controdiction.

Thus, the conclusion holds.