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My proof is this:

Let $f=u+iv$. We have $|f'|=0$, then $|u_x^2+v_x^2|\le |u_x|^2+|v_x|^2=0\Rightarrow u_x=0=v_x$. So $f'=u_x+iv_x=0$. As $f$ is holomorphic in $\mathbb C$, Cauchy–Riemann equations are true for $f$: $u_x=v_y$ and $u_y=-v_x$.

Taking the equation $f'=u_x+iv_x=v_y+iv_x=0$. Subtracting both equations: $u_x-v_y=0$. As we had $u_x=0$, then $v_y=0$.

Taking the same equation again: $f'=u_x+iv_x=u_x-iu_y=0$. Subtracting again: $i(u_x-u_y)=0\Rightarrow u_x=u_y=0$.

As we have all partials $u_x=u_y=v_x=v_y=0$, then $f$ is constant.

Do you think my proof is ok? Thanks!

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    Yes, it is correct. – 2017-01-02
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    More conceptual if $\gamma(t)$ is any curve the $f(\gamma(t))$ has zero derivative and so by then real case $f$ is constant along that curve, and since its true for all curves, $f$ is constant. – 2017-01-02
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    If you have that holomorphic $\implies$ analytic then you can just use the power series of $f$. – 2017-01-03

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