Assume we have an entire unbounded function $f(z)$.
My problem is the following contradiction I obtain:
We can conclude from Little Picard Theorem that the real part of $f(z)$ takes all real values (up to one). From the fact that the function is holomorphic everywhere I would conclude that the set $\{z\}$, where the $\Re f(z)$ is greater than zero is an open set, the same for the negative values. I also would think that at the boundary between the two $\Re f(z)=0$. But shouldn't then from identity theorem follow that $\Re f(z)=0$ everywhere (and accordingly our function is bounded)?
$\Re f$ is never holomorphic (unless $f$ is constant), therefore the fact that it vanishes on a non-discrete closed set means nothing: the essential fact is that it shouldn't vanish on any open set.
In fact, the set $(\Re f)^{-1}(0)$ is ideally a nice "curve" in $\Bbb R^2$; the set $(\Im f)^{-1}(0)$ is, likewise, another nice curve. And sice their itersection $f^{-1}(0)$ is very likely to be a discrete set, the identity theorem will stay put and behave.