The following is from Partial Differential Equations, Evans, p. 286~287.
Theorem. Assume $U$ is a bounded open subset of $\mathbb{R}^n$ and $\partial U$ is $C^1.$ Suppose $1\leq p < n.$ Then $W^{1,p}(U)$ is compactly embedded in $L^q(U)$ for each $1\leq q < p^*$, where $p^*$ denotes the Sobolev conjugate of $p$.
Here is a brief sketch of the proof.
(1) It is clear that $W^{1,p}\subset L^{p^*}(U)$ due to the Gagliardo-Nirenberg-Sobolev inequality. Since $U$ is bounded, the fact that $W^{1,p}(U)\subset L^q(U)$ for each $1\leq q < p^*$ is also clear. Therefore, it only remains to verify that the embedding is compact.
(2) Let $\{u_m\}$ denote a sequence that is bounded in $W^{1,p}(U)$. We must show that there is a subsequence $\{u_{m_j}\}$ that converges (to some limit) in the space $L^q(U)$. By extension we can assume WLOG that the $u_m$ are defined on all of $\mathbb{R}^n$, and supported compactly, with this support lying in some open set $V\supset U$. Moreover, the consequences of the extension theorem allow us to assume that $$\sup_m \|u_m\|_{W^{1,p}(V)} \leq C \sup_m \|u_m\|_{W^{1,p}(U)} < \infty.$$
(3) Consider the smoothed functions $u^\varepsilon_m := \eta_\varepsilon \star u_m$ first, where $\eta$ denotes the standard mollifier. By considering only small $\varepsilon>0$, we may assume that the $u^\varepsilon_m$ are all supported only in $V$.
(4) We first claim that $$u^\varepsilon_m \to u_m\text{ in }L^q(V)\text{ as }\varepsilon\to 0\text{ uniformly in $m$}.$$ To see this, assume first that $u_m$ is smooth. Then $$\vdots$$ (After some calculations) $$\int_V |u^\varepsilon_m (x) - u_m (x)| dx \leq \varepsilon \int_V |Du_m(z)| dz.$$ By approximation this estimate holds for general $u_m\in W^{1,p}(V).$
Alright, this is where I'm confused. We know that we can approximate any $u_m\in W^{1,p}(V)$ with functions that are smooth up to the boundary, say $\{u_m^r\}_r$ with $u_m^r\in C^\infty(\bar{V})$. This approximation is carried out in $W^{1,p}(V)$. But how exactly does this justify that the inequality above holds for general $u_m\in W^{1,p}(V)$? Is it because $u_m^r \to u_m$ and $Du_m^r \to Du_m$ in $L^p(V)$ and the fact that $L^p(V)\subset L^1(V)$ implies $u_m^r \to u_m$ and $Du_m^r \to Du_m$ in $L^1(V)$, so we can write \begin{align}\|u_m^\varepsilon - u_m\|_{L^1} &\leq \|u_m^\varepsilon - u_m^r\|_{L^1} + \|u_m^r - u_m\|_{L^1}\\ &\leq \varepsilon \|Du_m^r\|_{L^1} + \|u_m^r - u_m\|_{L^1} \\ &\to \varepsilon \|Du_m\|_{L^1}? \end{align} Or is there a more 'obvious' way to explain this?
I'd appreciate all help!
Not quite. $\newcommand{\eps}{\varepsilon}$ $u_m^\eps$ depends on $u_m$ by the definition $u_m^\eps := \eta_\eps \ast u_m$, so you can only apply the estimate with $(u_m^r)^\eps$, not with $u_m^\eps$. (That is, you need to replace all instances of $u_m$ by $u_m^r$.)
The statement follows by approximation (for fixed $\eps, m$) if \begin{align*} (u_m^r)^\eps \to u_m^\eps &\qquad \text{in $L^1$}, \\ (u_m^r) \to u_m &\qquad \text{in $L^1$} \quad \text{and} \\ (D u_m^r) \to D u_m &\qquad \text{in $L^1$}. \end{align*} Now you choose $(u_m^r)_r$ such that $u_m^r \to u_m$ in $W^{1, 1}$ which entails the last two convergence properties. The first one then follows by the second one (and a small computation).