This is a (slightly) more detailed version of the currently accepted answer

(For the sake of contradiction) Let $T_1 = (V, E_1)$ and $T_2 = (V, E_2)$ be two distinct MSTs of the graph $G = (V, E)$

Note that both have the same vertex set $V$ since both are *spanning* trees of $G$

Consider the set $E_{\Delta} = E_1 \triangle E_2$

Let $e = (u, v)$ be *the* edge in $E_{\Delta}$ having the least cost (or weight)

Note that since all costs are unique, and $E_{\Delta}$ is non-empty, $e$ must be unique.

Without loss of generality, assume $e \in E_1$

Now, there must be a path $P$, with $e \notin P$, in $T_2$ connecting $u$ and $v$, since trees are by definition, connected.

Note that *at least* one edge (say $e'$) that occurs in $P$ must not be in $E_1$, thus, $e' \in E_{\Delta}$

This is because, if $P \subset E_1$, $T_1$ will contain a cycle formed by the path $P$ and the edge $e$, this leads to a contradiction, since trees by definition are acyclic.

Note that by definition of $e$ and the fact that all costs are distinct, $\text{cost}(e) < \text{cost}(e')$

Now, consider $T_2' = (V, E_2')$, where $E_2' = (E_2 \backslash\{e'\})\cup\{e\}$, this has a strictly lesser total cost than $T_2$

As $T_2$ was a MST, this leads to a contradiction.