Picture below is from the 175th page of do Carmo's Riemannian Geometry. I don't know how to get the red line of first picture below.
$H^n=\{(x_1,...,x_n):x_n>0\}$ is the upper half-space.
What I try: assuming there is $q\in\partial H^n$ such that $h(q)\notin \partial H^n$, if $h(q)\in H^-=\{(x_1,...,x_n):x_n<0\}$, then there is sequence $q_n\rightarrow q$ such that $q_n\in H^n$ and $h(q_n)\rightarrow h(q)\in H^-$. Therefore, there must be some point $p\in H^n$ such that $h(p)\in H^-$, namely $h(p)\notin H^n$, this is contradict with $h$ take $H^n$ onto $H^n$.
But for the case $h(q)\in H^n$, I don't know how to prove it. I feel it will be contradict with conformal.
Definition of conformal
First of all, let's clean up do Carmo's treatment of the Luoiville Theorem. The correct statement of Theorem 5.2 is:
Suppose that $n\ge 3$, $U\subset S^n= R^n\cup \{\infty\}$ is a connected open subset and $f: U\to S^n$ is a conformal mapping. Then $f$ equals the restriction of a Moebius transformation of $S^n$.
[Here a Moebius transformation is a self-diffeomorphism of $S^n$ which is a composition of inversions. Equivalently, it is a composition of Euclidean isometries, Euclidean dilations (both extended to $S^n$) and inversions (at most one of each.]
With this formulation, the conformal map $h$ appearing in the proof of Theorem 7.5 you are referring to, is the restriction of a self-diffeomorphism of $S^n$. Hence, for each subset $W\subset S^n$, $h$ maps boundary points of $W$ to boundary points to $h(W)$.