This is in Section 2.8 of Gilbarg and Trudinger. I believe there are some inaccuracies in the proof supplied, and in any case I think there is a more straightforward proof.
Definition:
A $C^0(\Omega)$ function $u$ will be called subharmonic in $\Omega$ if for every ball $B\subset\subset \Omega$ and every function $h$ harmonic in $B$ satisfying $u\leq h$ on $\partial B$, we also have $u \leq h$ in $B$.
Statement:
(i) If $u$ is subharmonic in a domain $\Omega$, it satisfies the strong maximum principle; and if $v$ is superharmonic in a bounded domain $\Omega$ with $v\geq u$ on $\partial \Omega$, then either $v > u$ throughout $\Omega$ or $v\equiv u$.
The statement before the semicolon follows from $u\leq \bar{u}$ in any ball $B\subset\subset \Omega$, where $\bar{u}$ is the harmonic lifting of $u$ in $B$ (existence was proved as Theorem 2.6). In any case, it is a special case of the statement after the semicolon.
I won't comment on the author's proof, but instead I will present my own proof, which seems more straightforward and avoids the inaccuracies in the presented proof.
Lemma
Notice that if $u_1,u_2$ are subharmonic in a domain $\Omega$, then (by using Theorem 2.6 in every ball $B\subset\subset\Omega$) $u_1+u_2$ is subharmonic as well.
Proof
Thus $w = u-v$ is subharmonic in the given bounded domain $\Omega$. Furthermore $w\leq 0 $ on $\partial \Omega$. By the weak maximum principle, $w\leq 0$ in $\Omega$ (here is the only place we use the boundedness of $\Omega$).
If $w < 0$ fails to hold in $\Omega$, then (by the preceeding) there is a point $y\in\Omega$ where $w(y) = 0$. So by the strong maximum principle, $w\equiv 0$ in $\bar{\Omega}$.
Thus $v > u$ throughout $\Omega$ or $v\equiv u$. QED
Note: I might not know what I am doing, so please let me know if I am missing out on subtleties. When I start thinking there are mistakes in a book, it is a sure sign I am misunderstanding something. So please post your feedback, if you know something about this topic.