Let us fix $G$ and $T$, a minimum spanning tree (MST) of $G$ as in the question. Assume weights of all edges are distinct; we will come back to the general case later.
Since all weights are distinct, there is one-to-one correspondence between edges and their weights.
Let $e_i$ be the edge with weight $w_i$ and $e_i'$ be the edge with weight $w_i'$. Let $T_i$ be the graph with edges $\{e_1, \cdots, e_i\}$ for $i\le n-1$. The proof of Kruskal's algorithm in the case of $G$ proves, in fact, $G$ has a unique MST that will be found by Kruskal's algorithm, because all weights are distinct. Specifically, the case "Otherwise, $e$ is not in $T$" cannot happen.
Minimality Lemma. If an edge not in $T_i$ can be added to $T_i$ without creating a cycle, $e_{i+1}$ must not weigh more than that edge.
Proof. Kruskal's algorithm tries adding the edge of the least weight among the available edges repeatedly, thus including $e_1, e_2, \cdots$ to form $T$ in that order. Let us inspect the moment when Kruskal's algorithm adds $e_{i+1}$ to $T_i$. If $e$ is an edge that weigh less than $e_{i+1}$, it must have been tried to be added to $T_k$ for some $k\le i$. Either $e$ was added to $T_k\subseteq T_i$ or a cycle was created at that time, which means adding it to $T_i$ will create a cycle, too. Done.
For any graph $S$, let $p(S)=0$ if $e_1\not\in S$; otherwise let $P(S)$ be the largest index $k$ such that the first $k$ lightest edges in $S$ is the same as the first $k$ lightest edges in $T$. WLOG, let $T'$ be a counterexample such that $p(T')$ is the largest (among all spanning trees of $G$). Let $q=p(T')$.
Observation: $w_{q+1}\lt w'_{q+1}$.
Proof. $\{e_1, e_2\cdots, e_q\}\subseteq T'$. Consider $e_{q+1}'$, the next heavier edge in $T'$. The Minimality Lemma tells that $e_{q+1}$ does not weighs more than $e_{q+1}'$. If it weighs the same as $e_{q+1}'$, then it is $e_{q+1}'$, which contradicts to our assumption of $q$ being the largest. So it must weigh less than $e_{q+1}'$. Done.
Since $T'$ is a spanning tree, if we add $e_{q+1}\not\in T'$ to $T'$, we must have created a cycle. Since we do not have a cycle in $T_{q+1}$, there must be an edge $e'\not\in T_q$ in that cycle. The above observation tells us $e_{q+1}$ weighs less than $e'$.
Let $T''$ be $T'$ with $e'$ replaced by $e_{q+1}$. Since $e'$ and $e_{q+1}$ are on that same cycle, $T''$ is also a spanning tree. Since the only difference between $T'$ and $T''$ is one edge of $T'$ is replaced with an edge of smaller weight, $T''$ is also a counterexample. Since both the first $q+1$ lightest edges of $T''$ and that of $T$ are $\{w_1, w_2\cdots, w_q, w_{q+1}\}$, $p(T'')\ge q+1>q$. This contradicts the assumption that $q$ is the largest. Done.
How about the general cases when all weights are not distinct?
For any $\epsilon>0$ that is smaller than the minimum positive difference among the original weights, if we subtract $\epsilon/i$ weight from $e_i$ and subtract $\epsilon/(2dn)$ weight from other edges using different positive integer $d$ for different edges, all new weights will be different and, since edges of $T$ are reduced more, $T$ remains to be an MST.
Letting $\epsilon$ go to 0 to take the limit of the inequalities that we have proved in the cases of distinct weights, we will obtain the same inequalities in the general cases.
Interested readers may enjoy the following exercise.
Exercise. Generalize MST $T$ to minimum-weighted sub-forest.
Problem. Can we prove the above generalization directly?
