 Liouville rigidity from a bounded holomorphic embedding

Suppose $D\subseteq \mathbb{C}$ is open and admits an injective holomorphic map $\Phi :D\to \mathbb{C}$ whose image is bounded. Every entire function $f:\mathbb{C}\to D$ is constant: the function composition $\Phi \circ f$ is a bounded entire function, hence constant by Liouville theorem, and injectivity of $\Phi$ then forces $f$ to be constant. Surjectivity of $\Phi$ onto a disk is unnecessary. For example $z\mapsto z/(1-z)$ is a conformal map from $\mathbb{C}\setminus [0,1]$ onto $(\mathbb{C}\setminus [0,\infty ))\setminus \{-1\}$. A square-root branch with argument in $(0,2\pi )$ followed by the Cayley transform between the half-plane and disk maps this to the punctured unit disk. It therefore supplies a bounded injective holomorphic coordinate on the interval complement, giving the same rigidity even though that domain is not simply connected.