Prove that $f(\Bbb C)=\Bbb C$ I need some help with the following problem: 

Let $f:\Bbb C \to \Bbb C$ be continuous satisfying that $f(\Bbb C)$ is an open set and that $|f(z)| \to \infty$ as $z\to \infty$. Prove that $f(\Bbb C)=\Bbb C$. 

My idea on this one is to prove by contradiction and assume that $S=f(\Bbb C)\ne\Bbb C$ to get some contradiction with the given two properties of the function. But I have no idea on how to proceed next. 
Thanks in advance. 

By your condition, the image of $f$ is closed (why? take a convergent sequence of points in the image, and their preimages. The preimages do not diverge to infinity, by assumption, so they have a limit point, which, by continuity, goes to the limit you want). So, $f(\mathbb{C})$ is open (by assumption) and closed. So, since $\mathbb{C}$ is connected, the image is all of $\mathbb{C}.$
