Use induction to prove that radix sort works. Where does your proof need the assumption that the intermediate sort is stable?
We can take the following invariant:
At the beginning of the for loop, the array is sorted on the last $i - 1$ digits.
Initialization. The array is trivially sorted on the last 0 digits.
Maintenance. Let's assume that the array is sorted on the last $i - 1$ digits. After we sort on the $i$th digit, the array will be sorted on the last $i$ digits. It is obvious that elements with different digit in the $i$th position are ordered accordingly; in the case of the same $i$th digit, we still get a correct order, because we're using a stable sort and the elements were already sorted on the last $i - 1$ digits.
Termination. The loop terminates when $i = d + 1$. Since the invariant holds, we have the numbers sorted on $d$ digits.
We use the assumption in the maintenance explanation.