Does $W^{1,1}(\Omega)^*$ contain a countable separating set? I am trying to show that a weakly precompact subset $S$ of $W^{1,1}(\Omega)$ is metrizable. If I am not mistaken, it is enough to show that $W^{1,1}(\Omega)^*$ contain a countable separating set.

Is it the case that $W^{1,1}(\Omega)^*$ contains a countable separating set? 

Alternatively, if there is another approach to show that $(S,w)$ is metrizable I'd love to know about it as well.
 A: I think your idea in the comments just works.
First let's show that we can find a countable separating set $D \subset L^\infty(\Omega)$ for $L^1(\Omega)$. Define $\mathcal{A}$ to be the set of rectangles in $\Omega$ with rational endpoints and let 
$$D = \{ \mathbb{1}_A : A \in \mathcal{A}\} \subset L^\infty(\Omega).$$
Since $\mathcal{A}$ generates the usual Borel $\sigma$-algebra on $\Omega$, by a standard monotone class argument, 
$$\int_A f \, d \mu = 0 \text{ for all } A \in \mathcal{A} \implies \int_A f \, d \mu = 0 \text{ for all } A \in \mathcal{B}(\Omega).$$
In particular, this implies that if $f,g \in L^1(\Omega)$ and $\int_A f \, d \mu = \int_A g \, d \mu$ for every $A \in \mathcal{A}$ then $f = g$. This means that $D$ separates points in $L^1(\Omega)$. Clearly $D$ is also countable.
Now, recall that $W^{1,1}(\Omega) \hookrightarrow L^{1}(\Omega)$ and so by taking duals $L^\infty(\Omega) \hookrightarrow W^{1,1}(\Omega)^*$ and so we can regard $D$ as a subset of $W^{1,1}(\Omega)^*$. Then it is clear that $D$ is the desired countable separating set.
