There is a problem with your definition of removable discontinuity:

As f is continuous on all the reals, its derivative cannot have any removable discontinuities, i.e. discontinuities where the lefthand and righthand limits agree as x->c but f(c) does not exist

The function

f(x) = { x^2 for x!= 0

9 for x = 0

contains a removable discontinuity at x = 0, but f(0) exists.

With this in mind, if f'(x) were to have a removable discontinuity at x=0, it is possible that f'(0) may still exist.

So for this particular question, infinite removable discontinuities should be the only concern. I don't think you've shown that finite removable discontinuities are ruled out, but they won't cause problems for f'(0) not existing.

Despite this, your proof is fairly convincing. It's not very rigorously stated, but that's ok.

You can find a more immediate way of showing f'(0) exists using the Mean Value Theorem or L'Hospital's rule in combination with the definition of the derivative.