The solution is wrong: Consider $f(z) = z^n$ for (3)...
The rest is an application of the maximum modulus principle to $g$ and is correct. But we do need to have $f(0) = 0$ for $g$ to be holomorphic, as was already hinted at in the comments by Leonid Kovalev. If this additional information on $f$ is not given in the exercise statement, then all points stated there are wrong the first three points stated there are wrong: For (1) and (2) consider the constant function $f(z) = 1$.
(4) turns out to be right just from the assumption $|f(z)|\le 1$ alone. This follows from Cauchy's formula for the derivative. For any $0 we have
$f'(z) = \frac{1}{2\pi i}\oint_{|z| = 1-r} \frac{f(\zeta)}{(\zeta - z)^2} \, d\zeta$
In particular
$|f'(0)| \le \frac{1}{2\pi}\oint_{|z| = 1-r} \frac{|f(\zeta)|}{(1-r)^2} \, d\zeta \le \frac{1}{1-r}$
for all $0. Letting $r\to 0$ gives $|f'(0)| \le 1$.
Remark: In the case were $f(0) = 0$, you might want to check out the Schwarz Lemma.