How do I solve this Mean Value Theorem problem

The problem says:

If | f(w) - f(x) | < or = | w - x | for all values w and x, and f is a differentiable function, show that
-1 < or = f'(x) < or = 1 for all x values.

Every time I think I know how to do it, I get confused. Any help? Thanks.

Suppose that y and z are distinct.

Applying the Mean Value Theorem to f yields
f(y) - f(z) = f '(c) (y - z) for some c between y and z.

However, we know that |f(y) - f(z)| ≤ |y - z|.

Therefore, taking absolute values of the MVT relation yields
|f '(c) (y - z)| = |f(y) - f(z)| ≤ |y - z|
==> |f '(c)| |y - z| ≤ |y - z|
==> |f '(c)| ≤ 1.

Since y and z are arbitrary, we can conclude that |f '(x)| ≤ 1 for all x.
==> -1 ≤ f '(x) ≤ 1 for all x.

I hope this helps!