Consider an entire function $f$ such that $|Re f(z)|\ge |Im f(z)|, \forall z\in\mathbb{C}$ with $|z|\ge M$. Show that $f$ is constant. 
Consider an entire function $f$ such that $|Re f(z)|\ge |Im f(z)|,
 \forall z\in\mathbb{C}$ with $|z|\ge M\in\mathbb{R}$. Show that $f$ is constant.

My attempt:
The inequality has a geometric interpretation: For all points $z$ outside of $B(0,M)$ we can only reach points for which the real part is greater than or equal to the imaginary part. Select $w=a+bi$ with $a>M$ and $b>2a$, then the disk $B(w,M/2)$ is not reached by $f$, which implies that $f$ is constant.
Is this a good approach?
Thanks.

Your proof is incomplete. You did not prove that there is an open disk that doesn't intersect the range of $f$; all that you proved was that there is an open disk that doesn't intersect$$\left\{f(z)\,\middle|\,\bigl\lvert f(z)\bigr\rvert\geqslant M\right\}.\tag1$$However the range of $f$ is the union of $(1)$ with $f\left(\overline{B(0,M)}\right)$, which is a compact set, and therefore a bounded set. So, since the range of $f$ is a subset of$$\left\{z\in\mathbb Z\,\middle|\,\lvert\operatorname{Re}z\rvert\geqslant\lvert\operatorname{Im}z\rvert\right\}\cup f\left(\overline{B(0,M)}\right)$$which is a closed set but which is not the whole complex plane, the range of $f$ cannot possibly be a desnse subset of $\mathbb C$.
