By Rellich theorem we knoow that for a open, bounded Lipschitz domain $\Omega$ in $\mathbb{R}^n$ we have : $H^{1}(\Omega)$ is compactly embedded in $L^{2}(\Omega)$.
Now, I want to prove that : $L^{2}(\Omega)$ is compactly embedded in $H^{-1}(\Omega)$.
Here's a proof using general functional analysis machinery.
Schauder's theorem (you can find a discussion of this in the top answer here) tells us that if $X$ and $Y$ are Banach spaces, then $T: X \to Y$ is compact if and only if $T^\ast : Y^\ast \to X^\ast$ is compact. Rellich tells us that the inclusion map $I: H^1_0 \to L^2$ is compact, so $I^\ast : (L^2)^\ast \to (H^1_0)^\ast = H^{-1}$ is also compact by Schauder's theorem. Since we can canonically identify $L^2 = (L^2)^\ast$, we're done as soon as we unravel the action of $I^\ast$. For $f \in H^1_0$ and $g \in L^2$, $$ \langle I^\ast g,f \rangle_{H^{-1}} = \langle If , g \rangle_{(L^2)^\ast} = \int_\Omega fg, $$ and so $I^\ast$ is the usual injection of $L^2$ into $H^{-1}$, so we're done.
