Question: can we claim that $W^{1,p}(\Omega) \subset L^1(\Omega) \subset W^{-1,p}(\Omega)$?

Let $$\Omega$$ be a compact manifold in $$\mathbb R^2$$. For $$1 \leq p \lt 4/3$$ can we claim that

$$W^{1,p}(\Omega) \subset L^1(\Omega) \subset W^{-1,p}(\Omega)$$

with the first inclusion being compact and the second one being continuous?

Note that $$W^{-1,p}(\Omega)$$ is identified with $$W^{-1,(p')'}(\Omega)$$ which is the dual space of $$W^{1,p'}(\Omega)$$

MOTIVATION: I want to use an Aubin-Lions-type lemma without the reflexivity assumption but I'm not sure about the mentioned inclusions.

Any help or hint is much appreciated.

For the first compact embedding: The Rellich-Kondrachov theorem guarantees that in dimension $$d$$ and for $$p we have $$W^{1,p}\subset\subset L^q$$ for all $$1\leq q. Regardless of the specific value of $$(4/3)^*=4$$ in dimension 2, this always holds for $$q=1$$ so you're good.
As for the second part of your problem: this sort of question is better though of by "dualizing", i-e asking whether one has dual embedding $$W^{1,p'}\subset L^\infty\quad ??$$ for $$p>\frac 43$$. If so then clearly your embedding holds, and in fact the $$W^{-1,p}$$ action of an $$L^1$$ function will be just the usual Lebesgue integration (because in that case $$|\int u\varphi|\leq \|u\|_1\|u\|_\infty\leq \|u\|_1\|\varphi\|_{W^{1,p'}}$$ will be a $$W^{1,p'}$$-continuous linear form). Here we have explicitly $$p'>(4/3)'=\frac{\frac{4}{3}}{\frac{4}{3}-1}= 4$$ for all $$p<4/3$$. In dimension $$d=2$$ this means that you're in the "supercritical" regime $$p'>d$$ for the Sobolev embedding, therefore by Morrey's inequality you have indeed $$W^{1,p}\subset L^\infty$$ (actually $$W^{1,p}$$ embeds into a much better Hölder space with exponent depending on $$p$$ and the dimension $$2$$). So the answer is yes to that second part too.