Problem:
Let $f$ be an entire function such that $|f(z)|\neq1 \: \: \: \: \forall z\in \mathbb{C} $ then prove $f$ is constant
Solution: (the reason that I am asking here $\rightarrow $) does that mean that $|f(z)|\ge 1\: \: \forall z\in \mathbb{C}$ or $|f(z)|\leq 1\: \: \forall z\in \mathbb{C}$ ?
If that's the case I think I know how to solve it, but I can't prove that's the only 2 cases.
If $|f(z)|\ge 1\: \: \forall z\in \mathbb{C}$ that means $f(z)\ne 0$ and we consider the function $g(z)=\frac{1}{f}$, but $|g|<1$ which means from Liouville theorem $g$ is constant therefore $f$ is constant.
For the case that $|f(z)|\leq 1\: \: \forall z\in \mathbb{C}$ $f$ is constant for the same reason.
Is my solution correct?
No, it does not mean that. It means that for each $z\in\Bbb C$, either $|f(z)|>1$, or $|f(z)|<1$.
However, since $f$ is entire, it is continuous. So, since $\Bbb C$ is connected, $f(\Bbb C)$ is connected too. Therefore, either $f(\Bbb C)\subset D_1(0)$, or $f(\Bbb C)\subset\Bbb C\setminus\overline{D_1(0)}$. And then you can apply your argument.