Are edges in a minimum spanning tree not heavier than respective edges in another spanning tree?
Here are some hints:


Flesh out your counterexample with quantifiers properly: "There exists a graph $G$, and a minimum spanning tree $T$ for $G$ with edge weights $w_1 \le w_2 \le \dots \le w_{n-1}$, and another spanning tree $T'$ of $G$ with edge weights $w'_1 \le w'_2 \le \dots \le w'_{n-1}$, and an integer $i$ obeying $1 \le i \le n-1$, such that $w'_i &lt; w_i$".  In a situation like this, from the set of all such counterexamples ($(G, T, T', i)$ tuples), you are free to choose one having minimum possible $i$, and it's nearly always useful (and never harmful) to do so.
"I tried adding it to the tree $T$" -- you must first show that it is not already in $T$.  Here you can use minimality of $i$ for edges to the left, and the increasing weight order for edges to the right.
Let's add it to $T$, to make a graph I'll call $G_2$.  What noticeable feature does $G_2$ contain?
What simple modifications could you make to that feature in $G_2$ to turn it back into a tree?  Hint: With this particular feature, there will always be at least 3 ways to do this, and there could be as many as $n$.
Can you choose an option in the previous step that will lead to a tree that is strictly lighter than $T$, and thus a contradiction (since we assumed $T$ to be minimal)?