As mentioned by BlueRaja you can prove this with only three nodes.
Take 3 nodes (A,B,C) with the following edge weights:
A->B = 1
B->C = 2
A->C = 3
Now if you take A as source vertex and try to find the shortest paths you can get two different trees.
1- The edge AC and AB are present. Cost to C=3 and Cost to B=1.
2- The edge AB,BC are present. Cost to C=3 and Cost to B=1.
As you can see all edge weights are distinct yet we are getting two different shortest spanning trees.
The basic idea is that:
Let C be the least cost of reaching a particular node in a spanning tree. Then it may be possible to find another path to C with the same cost if the sum of edge weights constituting the path is equal to C.