Let $f$ be holomorphic on $\Omega\subset\Bbb C$, and $\Omega$ is open and connected. If $|f(z)|$ is constant it is known that $f$ is constant on $\Omega$
The proof of this fact says that if $f$ is not constant then $f(\Omega)$ is open as the image of an open by a holomorphic function. But a circle of radius $c\in\Bbb R_+$ is not open and $f(\Omega)$ wouldn't be open either as a subset of that circle, which contradicts the first sentence.
Question 1
Where is the connectedness of $\Omega$ used in the above proof?
Question 2
Why should $\Omega$ have any of the two constraints? Is the following proof invalid?
writing $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$ since $f$ is holomorphic the Cauchy-Riemann equations hold: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}~~\text{ and }~~\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ $\forall (x+iy)\in\Omega\ ~|f(x+iy)|=\sqrt{u^2(x,y)+v^2(x,y)}=c\in\Bbb R_+$
$\implies\frac{\partial }{\partial x}[\sqrt{u^2(x,y)+v^2(x,y)}]=\frac{\partial }{\partial y}[\sqrt{u^2(x,y)+v^2(x,y)}]=0$
$\iff (u_x+v_x)(u^2+v^2)^{-{1\over2}}=(u_y+v_y)(u^2+v^2)^{-{1\over2}}=0$
If $u^2+v^2=|f|^2=0\forall z\in\Omega$ then $f=0$ and we're done
Let's suppose $u^2+v^2\ne0$ then we can divide by $(u^2+v^2)^{-{1\over2}}$ it on both sides to get: $$u_x+v_x=u_y+v_y=0$$ by the C-R equations $$u_x-u_y=u_y+u_x\implies u_y=-v_x=0$$ substituting $v_x=0$ into $u_x+v_x=0$ $$u_x=0\implies v_y=0$$
All the partial derivatives of $u$ and $v$ are zero and since $f'(z)=u_x(x,y)+iv_x(x,y)=0$ we get that $f$ is constant.