Let
$g(z):=\frac{f(z)-f(0)}z$ for $z\neq 0$ and $g(0)=f'(0)$. It's still a holomorphic function, and
$$|g(z)|\leq \frac{2c(1+|z|^{1/2})}{|z|},$$
hence $g$ is bounded by $1$ for $|z|\geq R$ for some $R$. Since $g$ is continuous on $\overline{B(0,R)}$, $g$ is bounded on this set. By Liouville's theorem, $g$ is constant, hence $f(z)-f(0)=cz$ for some $c\in\Bbb C$. Using this in the initial assumption, we get that $c=0$ and $f$ is constant.
An other approach is to use Cauchy's integral formula directly:
$$f'(z)=\frac 1{2\pi i}\int_{C(z,R)}\frac{f(\xi)}{(z-\xi)^2}d\xi,$$
hence for $R\geq 1$,
$$|f'(z)|\leq \frac{c(1+\sqrt R)}{R^2}R=\frac c{\sqrt R}.$$
We get $f'(z)=0$ for all $z\in\Bbb C$ (connected) hence $f$ is constant.