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.
Thanks in advance!
 A: The answer is yes for both.

For the first compact embedding: The Rellich-Kondrachov theorem guarantees that in dimension $d$ and for $p<d$ we have $W^{1,p}\subset\subset L^q$ for all $1\leq q<p^*=\frac{dp}{d-p}$. 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.
