Suppose $f$ is continuous on $S=\{|z-z_0|<\rho\}$. Then we can find some $0<\rho'<\rho$ such that on the circle $\{|z-z_0|=\rho'\}$ we get $|f(z_0)|\leq 2|f(z)|$.
This is the part that I don't understand from the proof of Hurwitz' theorem. How continuity implies it?
Since $f$ is continuous, there is some $\delta>0$ such that$$|z-z_0|<\delta\implies\bigl|f(z)-f(z_0)\bigr|<\bigl|f(z_0)\bigr|.$$Take $\rho'=\dfrac\delta2$. Then, if $|z-z_0|=\rho'$,\begin{align}\bigl|f(z)\bigr|&=\bigl|f(z)-f(z_0)+f(z_0)\bigr|\\&\leqslant\bigl|f(z)-f(z_0)\bigr|+\bigl|f(z_0)\bigr|\\&\leqslant2\bigl|f(z_0)\bigr|.\end{align}