The proof begins by considering the open set $D_{\varepsilon} = \{ x \in \Omega : u(x) > v(x) + \varepsilon \}$ for some $\varepsilon > 0. $ Then it says that $D_{\varepsilon} = \emptyset$ or $D_{\epsilon}$ is compactly contained in $\Omega.$ Clearly if $D_{\epsilon} = \emptyset$ then there is nothing to prove. But how do I show that if $D_{\varepsilon} \neq \emptyset$, then $D_{\varepsilon}$ is compactly contained in $\Omega.$ Am I using the assumption that $\limsup u(x) \leq \liminf v(x)?$
Let $K$ be the closure of $D_\epsilon$. $K$ is compact because it is bounded and closed. If we can show that $K \subset \Omega$ then $$ D_\epsilon \subset K \subset \Omega $$ so that $D_\epsilon$ is compactly contained in $\Omega$.
We surely have that $K \subset \overline \Omega$. If $K \not\subset \Omega$ then $$ K \cap \partial \Omega = \partial D_\epsilon \cap \partial \Omega $$ is not empty, so that there is a sequence $(x_n)$ with $x_n \in D_\epsilon$ which converges to a point $\zeta \in \partial \Omega$. But then $$ \limsup_{x \to \zeta} u(x)- \liminf_{x \to \zeta} v(x) \ge \limsup_{n \to \infty} u(x_n) - \liminf_{n \to \infty} v(x_n) \ge \limsup_{n \to \infty} \bigl(u(x_n) -v(x_n)\bigr)\ge \epsilon $$ is a contradiction to the given boundary condition.
Note that specific properties of the functions $u$ and $v$ (such as being $p$-harmonic) are not needed for this conclusion, only the boundary condition.