Step 1: Instead of relying on a single example, exhibit an entire family of counterexamples so the failure of statement (C) is unmistakable: for any real constant $c>0$, let $f_c(z)=e^{cz}$, which is entire.
Step 2: For $z=x+iy$ with $\operatorname{Re}(z)=x\le0$ and $c>0$, $|f_c(z)|=e^{\operatorname{Re}(cz)}=e^{cx}\le e^{0}=1$, since $cx\le0$. So every $f_c$ is bounded (by $1$) on the half-plane $\{\operatorname{Re}(z)\le0\}$.
Step 3: Yet $f_c'(z)=ce^{cz}\neq0$ for every $c>0$, so $f_c$ is non-constant. This produces infinitely many non-constant entire functions satisfying the boundedness hypothesis in (C), so the implication bounded on $\{\operatorname{Re}(z)\le0\}\Rightarrow$ constant cannot hold; statement (C) is false.
Step 4: This also explains, by contrast, why (D) is correct: boundedness of $|f|$ on a mere half-plane is achievable by exponentials because $\operatorname{Re}(cz)$ can be forced non-positive there, but boundedness of $\operatorname{Re}(f)$ over the ENTIRE plane is a much stronger global restriction that only constants can satisfy, by Liouville's theorem applied to $e^{f}$.\[\boxed{\text{Option (C) is false}}\]