This approach does not work.
Assume your weights are 1,2,2,5. Then the minimal difference is 0, which can only be attained by putting the first three stones in one suck, and the last stone in the other.
Your algorithm, however, will divide the weights such that the difference is 4.
Notice, however, that this problem can be transposed. Assume you have a single sack, but for each weight you can choose if it's either positive or negative (laws of physics be damned). The weight in the new sack is the difference of weights between the original sacks, where the sign decides which of the original sacks the weight goes to.
Taking it a step further, this translates to the following problem:
Given a list of positive numbers, find a way to sign them such that the absolute value of the sum is minimal.
This problem can be solved with dynamic programming with runtime complexity of $o(n^2)$, where $n$ is the amount of numbers (under the relaxing assumption that addition/subtraction of two numbers has constant complexity).
I mistakenly thought you sort them ascending and not descending.
For the descending case your solution seems to work (and I like it better than my proposed solution).
To prove it assume that there's an optimal solution which gets a lower weight difference than your result. Choose the sack with the biggest rock to be S, and let i be the maximal index such that the rock m[i] was placed differently than your algorithm would given the current state of the sacks just before it was placed.
If m[i] is the last stone and the sacks aren't of equal weight then obviously the solution is not optimal (the last stone was put on the heavier pile).
If m[i] is the last stone and the sacks are of equal weight then, since this is the lightest stone, then your algorithm would've reach the same difference.
If m[i] is not the last stone, then had you put it on the other stack, your algorithm would produce a better result, contradicting the given solution being optimal.