Here's an answer that spits out a number in reverse, instead of reversing a string, by repeatedly dividing it by 10 and getting the remainder each time:

num = int(input("Enter a number: "))

while num > 0:
    num, remainder = divmod(num, 10)
    print remainder,

Oh and I didn't read the requirements carefully either! It has to be a for loop. Tsk.

from math import ceil, log10

num = int(input("Enter a number: "))

for i in range(int(ceil(math.log10(num)))): # => how many digits in the number
    num, remainder = divmod(num, 10)
    print remainder,

Use reversed() function (efficient since range implements __reversed__):

reversed(range(10))

It's much more meaningful.

Update: list cast

If you want it to be a list (as @btk pointed out):

list(reversed(range(10)))

Update: range-only solution

If you want to use only range to achieve the same result, you can use all its parameters. range(start, stop, step)

For example, to generate a list [3, 2, 1, 0], you can use the following:

range(3, -1, -1)

It may be less intuitive, but it works the same with less text. This answer by @Wolf indicates this approach is slightly faster than reversed.