Prove that there is no entire function such that for every $w$ in the complex plane $f^{-1}(w)$ consists of exactly 2 points. So I came across the following problem.
Prove that there is no entire function such that for every $w$ in the complex plane $f^{-1}(w)$ consists of exactly 2 points. 
So here is what I was thinking. Let $f^{-1}(w)=\{z_1, z_2\}$. Now consider $f: \mathbb{C} \setminus \{z_1, z_2\}  \rightarrow \mathbb{C} \setminus \{w\}$, the restriction. Now consider the open mapping property of holomorphic functions. Given any point $x\in \mathbb{C} \setminus \{w\}$ we can pick two sufficiently small open balls around the preimage points so that they map to an open neighborhoods around $x$. . Take their intersection, and take an open ball around $x$. 
Now, I would like to say taking the preimage of this open neighborhood gives me two disjoint open subsets homeomorphic to a ball, and so I would have a 2 sheeted covering. Coverings induce injections on fundamental groups, but the fundamental group of the twice punctured plane is $\mathbb{Z}\ast \mathbb{Z}$ which is non-abelian, while that of the once punctured plane is just $\mathbb{Z}$ yielding a contradiction. 
So, does this seem correct? And is there a more clever way of doing this without resorting to covering space ideas?
Thanks for any help!

Here's a proof that only uses tools from complex analysis. Every entire function such that the preimage of every point is bounded is a polynomial. A two-point set is of course bounded, so your function $f$ would have to be a polynomial. For a small enough perturbation $\epsilon$, the roots of $f(z) + \epsilon$ are simple; but $-\epsilon$ only has two preimages by $f$, so $f$ has degree $2$.
Let $f(z) = az^2+bz+c$. For $d \in \mathbb{C}$, the discriminant of $f(z) + d$ is $b^2 - 4a(c+d)$, so if you let $d = \frac{b^2}{4a} - c$, you find that $f(z) + d$ has a double root. In other words $f(z) = -d$ has a unique solution, a contradiction.
