Prove that $f(\mathbb{C}) = \mathbb{C}$ for a continuous function $f: \mathbb{C} \rightarrow \mathbb{C}$ with the following properties. The properties are as follows:


*

*$|f(z)| \rightarrow \infty$ as $|z| \rightarrow \infty$

*$f( \mathbb{C})$ is an open set


I start off by supposing that $f( \mathbb{C}) \neq \mathbb{C}$ in order to get a contradiction, i.e. there exists some $c \in \mathbb{C} \setminus f(\mathbb{C})$, but I'm not sure where to go from here, so I would appreciate any insight.
Can the following theorem be made use of?


*

*If G and D are plane domains (non-empty, open, connected sets) such
that $G \subset D$, then $\partial G$ $\cap $ $D \neq \emptyset$, unless
$G=D$.

 A: If $f(\mathbb C)\neq\mathbb C$ then the boundary of $f(\mathbb C)$ is not empty. Let $w\in\partial f(\mathbb C)$. Let $(z_n)_{n\in\mathbb N}$ be a sequence of elements of $\mathbb C$ such that $\lim_{n\to\infty}f(z_n)=w$. There are two possibilities now:


*

*The sequence is bounded. Then it has a subsequence $(z_{n_k})_{k\in\mathbb N}$ which converges to some $z$. But then$$w=f\left(\lim_{k\to\infty}z_{n_k}\right)=f(z),$$which is impossible, since $w$ would then be an interior point of $f(\mathbb C)$ (since $f(\mathbb C)$ is open).

*The sequence is unbounded. Then it has a subsequence $(z_{n_k})_{k\in\mathbb N}$ such that $\lim_{k\to\infty}\lvert z_{n_k}\rvert=\infty$. But then $\lim_{k\in\infty}\bigl\lvert f(z_{n_k})\bigr\rvert=\infty$ too, whereas this limit should be $\lvert w\rvert$.

A: I will offer a slight variation (since you already have an answer), if we can let $f$ be holomorphic instead of just continuous
We can assume $f$ is not constant, or else $f(\mathbb{C})=\mathbb{C}$ and we are done.  
Assme by way of contradiciton that $f(\mathbb{C})\neq\mathbb{C}$, so that there exists $c \in \mathbb{C} \setminus f(\mathbb{C})$.  Then we can define a continuous function
$$g(z) = \frac{1}{f(z)-c}\qquad z\in\mathbb{C}$$
such that $g(z) \to 0$ for $|z| \to \infty$ and is defined for all $z$. This means $g$ is bounded hence constant by Liouville's theorem.  Thus $f$ is constant, a contradiction
