= 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 disk>
$$
\Phi_1(z)=\frac{z-i}{z+i}.
$$
Its denominator is nonzero there, and $|z+i|^2-|z-i|^2=4\operatorname{Im}z>0$, 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 branch
$$
S(z)=\sqrt{|z|}\,e^{i\operatorname{Arg}z/2},\qquad0<\operatorname{Arg}z<2\pi.
$$
This 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 $1/(2S(z))\neq0$. Therefore
$$
\Phi_2(z)=\frac{S(z)-i}{S(z)+i}
$$
is an <injective> <conformal map> of $U_2$ onto the <unit disk>.
For the complement of the closed interval, first use
$$
M(z)=\frac{z}{1-z}.
$$
Its only finite pole is the excluded point one, and it is <injective> with <derivative> $1/(1-z)^2\neq0$. If $M(z)=t\geq0$, solving gives $z=t/(1+t)\in[0,1)$, which is excluded. Hence $M(U_3)\subset U_2$, and
$$
\Phi_3(z)=\Phi_2\left(\frac{z}{1-z}\right)
$$
is 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 gives
$$
\boxed{f:\mathbb C\to U_j\text{ holomorphic}\Longrightarrow f\text{ constant},\qquad j=1,2,3.}
$$
This illustrates <Liouville rigidity from a bounded holomorphic embedding>: an explicit <injective> <bounded> coordinate is enough.
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