 Solution

It suffices to construct an injective holomorphic map ${\Phi }_j:U_j\to \mathbb{D}$ for each domain. Then ${\Phi }_j\circ f$ is a bounded entire function and is constant by Liouville's theorem; injectivity of ${\Phi }_j$ forces $f$ to be constant.

For the upper half-plane, use the Cayley transform between the half-plane and diskIts denominator is nonzero there, and $|z+i{|}^2-|z-i{|}^2=4\mathrm{Im}z>0$, so its image lies in the unit disk. The determinant is nonzero, hence it is injective and conformal.

On the plane slit along the nonnegative real axis, choose the branchThis is the holomorphic square root obtained from the corresponding logarithm branch. Its argument lies in $(0,\pi )$, so it maps $U_2$ bijectively onto the upper half-plane, with inverse $w\mapsto w^2$. Its derivative is $1/(2S(z))\ne 0$. Thereforeis an injective conformal map of $U_2$ onto the unit disk.

For the complement of the closed interval, first useIts only finite pole is the excluded point one, and it is injective with derivative $1/(1-z{)}^2\ne 0$. If $M(z)=t\ge 0$, solving gives $z=t/(1+t)\in [0,1)$, which is excluded. Hence $M(U_3)\subset U_2$, andis the required injective conformal map into the disk. More precisely, $M(U_3)=U_2\setminus \{-1\}$ and ${\Phi }_3(U_3)=\mathbb{D}\setminus \{0\}$, since $S(-1)=i$. No claim that the interval complement is conformally equivalent to the whole disk is required.

Applying Liouville to the three bounded compositions givesThis illustrates Liouville rigidity from a bounded holomorphic embedding: an explicit injective bounded coordinate is enough.