complex analysis topology problem $\underline{\textbf{Given:}}$ 
(1) $\;f\,$ is a continuous function. 
(2) $\;f \,: \mathbb{C} \to \mathbb{C}.$ 
(3) $\;|f(z)| \to \infty\;$ as $\;|z| \to \infty.$ 
(4) $\;f(\mathbb{C})\;$ is an open set. 
$\underline{\textbf{To Prove:}}$ 
(5) $\;f(\mathbb{C}) = \mathbb{C}.$
$\underline{\textbf{My Request:}}$ 
I provide context and analysis below.  I welcome any hint, guide, or proof of
the problem.  It will help me if you keep your responses basic (i.e. consistent
with the context).
$\underline{\textbf{Context:}}$ 
I am recreationally self-studying "An Introduction to Complex Function Theory" : 
Bruce Palka : 1991.  My query represents the very last exercise from chapter 2
of this book (i.e. exercise 5.37).
Chapter 1 of this book focuses on the complex number system and chapter 2 
focuses on the rudiments of plane topology.  Within chapter 2 are definitions,
theorems and (end-of-chapter) exercises that focus on 
A. disks, open/closed sets, boundary points, sequences, convergence, accumulation points 
B. continuity, limits 
C. connected/disconnected sets, domains, components of open sets 
D. bounded sets and sequences, Cauchy sequences, compact sets, uniform continuity.
A set is connected $\;\Leftrightarrow\;$ the set is not disconnected. 
A set $\,S\,$ is disconnected $\;\Leftrightarrow\; \exists \;$ disjoint open sets $\;U, \,V \;\ni$ 
$(S \bigcap U) \neq \phi \neq (S \bigcap V)\;$ and $\;S \;\subseteq \;(U \bigcup V).$
$b\,$ is a boundary point of a set $\,S\; \equiv \;$ 
$\forall \;r > 0, \Delta(0,r)\;$ contains at least one element from $\,S\,$ 
and one element that is not in $\,S.$
I interpret premise (3) above as: 
$\forall \;\epsilon > 0 \;\exists \;r > 0 \;\ni \;|z| \geq r \;\Rightarrow \;|f(z)| > \epsilon.$ 
Please let me know if you disagree with my interpretation.
I completed all prior exercises in chapter 2.  The problem (exercise 5.37) provides
a hint: 
Assume $\;G = f(\mathbb{C}) \neq \mathbb{C}\;$ and use exercise 5.25 from chapter
2 to get a contradiction. 
Exercise 5.25 : 
If $\;G, \,D \;$ are domains (i.e. open, connected sets) $\;\ni$ 
$G \subseteq D\;$ and $\;G \neq D,\;$ then 
$\;\partial G \;\bigcap D \;\neq \;\phi$ 
(i.e. $\,D\,$ contains at least one of the boundary points of $\,G$).
$\underline{\textbf{My Analysis:}}$ 
First of all, let $\;h(z) = |f(z)|.$ 
From the theorems in chapter 2, I know that $\;h(z)$ is a continuous function, 
and that $\;h(\mathbb{C})\;$ is therefore an unbounded, open, and connected set. 
This means that $\;\forall \;r > 0, \;\exists \;z \,\in \,\mathbb{C} \;\ni
\;|f(z)| = r.$
Second of all, following the hint, and assuming that $\;G \neq \mathbb{C},\;$ 
$G$ has a boundary point $\,b \;\ni \;b \,\not\in \,G.$ 
Let $\;r \equiv |b|,\;$ and $\;K(0,r) \equiv \{ \,z \,: \;|z| = r \,\}.$ 
Let $\;K_1 \equiv G \bigcap K(0,r)\;, K_2 \equiv [K(0,r) \sim K_1].$
From previous exercises in chapter 2, I know that both $\;K(0,r)\;$ and $\;\overline{G}\;$
are connected sets. 
However, I do not (as of yet) know whether any of $\;\partial G, \;K_1, \;$ or $\;K_2 \;$ are connected
 sets. 
Further, even if I did know this, I don't see how it would help.
Since $\,b\,$ is a boundary point of $\,G,\,$ both $\,G\,$ and $\,(\mathbb{C} \sim G)\,$ 
contain elements arbitrarily close to $\,b.$ 
I don't see how this helps either.
$\underline{\textbf{Addendum-1 Reaction to Qiyu Wen's answer:}}$ 
I followed almost all of the answer, but I have some questions. 
In response to your paragraph:
"I just want to mention a gap in your interpretation: ... so 0 is what we try to capture." 
Q1:
What does $\;\mathbb{C^*}\;$ refer to?
Q2:
By "gap in your interpretation," are you referring to my presumption that
$\forall \;\epsilon > 0 \;\exists \;r > 0 \;\ni \;|z| \geq r \;\Rightarrow \;|f(z)| > \epsilon$ 
or are you referring to my analysis that concluded that 
$\;\forall \;r > 0, \;\exists \;z \,\in \,\mathbb{C} \;\ni \;|f(z)| = r?$
If the former, then how (else) does one interpret 
$\;|f(z)| \to \infty\;$ as $\;|z| \to \infty?$ 
If the latter, I know that since $\,f(z)$ is continuous, so is $\,h(z) = |f(z)|.$ 
Further, I know that since $\,\mathbb{C}\,$ is a connected set, so is $\,h(\mathbb{C}).$ 
I also know, from premise (3), that $\,h(\mathbb{C})\,$ is unbounded. 
However, I now realize that I do not immediately know that $\,{0} \,\in \,h(\mathbb{C}).$ 
Is this what you are referring to?
Q3:
In your statement "Since $\,f(C)\,$ is open, there exists an open disk cantered at $\,w$", 
do you intend ...cantered at $\,w_0$?
Q4:
In the paragraph that begins: "Suppose, for contradiction, that $\,w_0 \neq 0$ ..." 
I agree that both $\;f(\mathbb{C})\;$ and $\;h(\mathbb{C})\;$ are open sets, 
since both $\,f\,$ and $\,h = |f|\,$ are continuous. 
I need to be clear about your reasoning in this paragraph. 
Are you saying that since $\;h(\mathbb{C})\;$ is an open set, 
$\exists \;\delta > 0 \;\ni \;\Delta(|w_0|, \delta) \,\subseteq\,h(\mathbb{C})?$
Q5:
At the end of your first paragraph, you say "...Note that $\,w_0 < M.$" 
Why?  I agree that $\,|f|\,$ has a minimum at some point $\,z_0\,$ 
on the closed disk $\,D,\,$ but I don't understand why $\,|f(z_0)|\,$ must be
$\,< M.$ 
If I'm right that this is unclear, I think that the analysis is immediately remedied 
by simply taking any random $\,z_1 \,\in \,\mathbb{C}\,$ and then choosing 
$\,M > |f(z_1)|.$ 
What do you think?
Q6:
If I'm not mistaken, you ignored the hint given by the book. 
Is there a relation between your approach and the book's hint?
$\underline{\textbf{Addendum-2 Reaction to Qiyu Wen's response to my questions:}}$ 
Way cool!
"...I'm working with $\,f(C),\,$ which I know is open, not $\,h(C).$" 
"You pick a non-zero complex number, then in any ball about it"  
"you can find another complex number closer to 0..."
It's a critical point.
When I first saw your approach, I was blind to the above idea.
When you emphasized it, it sunk in.  Thanks
"...I didn't realize there's a hint. See my edited answer." 
Again, thanks.  The hint-approach, which is probably the intended answer, is
something that I might eventually have stumbled into.  Re your initial approach,
there's no way that I would have thought of that on my own.
 A: We first prove that $f(z)=0$ for some $z\in\mathbb{C}$. Pick $M$ in the range of $|f|$. Since $|f(z)|\to+\infty$ as $|z|\to+\infty$, there is a closed disk $D$ such that $|f(z)|>M$ for all $z\notin D$. By continuity of $f$ and compactness of $D$, $|f|$ has a minimum on $D$ at some $z_0\in D$. Put $w_0=f(z_0)$. Note that $|w_0|\leq M$. Hence $|f(z_0)|$ is a minimum of $|f|$ on $\mathbb{C}$.
Suppose, for contradiction, that $w_0\neq 0$. Since $f(\mathbb{C})$ is open, there exists an open disk cantered at $w_0$ and contained in the range of $f$, from which we can pick $w$ such that $|w|<|w_0|$, a contradiction.
Now let $w$ be any complex number and put $g(z)=f(z)-w$. Then $g$ also has the properties listed, so $g(z)=0$ for some complex number $z$. Hence $f$ is surjective.
I just want to mention a gap in your interpretation: you do not immediately know that $|f(\mathbb{C})|$ is open in $\mathbb{R}$, because $z\mapsto |z|$ is not an open map from $\mathbb{C}$ to $\mathbb{R}$. However, it is open on $\mathbb{C}^*$, so $0$ is what we try to capture.
This is how you can use the hint: Suppose that $f(\mathbb{C})\neq\mathbb{C}$. Let $w_0$ be a boundary point of $f(\mathbb{C})$. Let $(w_k)$ be a sequence in $f(\mathbb{C})$ that converges to $w_0$. Then you obtain a sequence $(z_k)$ such that $f(z_k)=w_k$. Use $\lim|f(z)|\to+\infty$ to show that $(z_k)$ is a sequence in a compact set, hence admits a convergent subsequence with limit $z_0$. Then by continuity of $f$ we have $f(z_0)=w_0$, contradicting $w_0\notin f(\mathbb{C})$.
These approaches are not that different. One way or another you want to show that $f(\mathbb{C})$ cannot have a boundary point.
