$L^1$ norm equivalent to weak topology of $W^{1,1}$?

By Rellich-Kondrachov the inclusion map from $W^{1,1}$ to $L^1$ is compact (with respect to the norm topologies). And a compact operator is weak-to-norm sequentially continuous (standard exercise, use a double subsequence trick and Hahn-Banach). So if we are right about $S$ being weakly metrizable, then it is true that the inclusion map from $(S,w)$ into $L^1$ is continuous and hence a homeomorphism onto its image.

(I got a little worried about the weak metrizability when Google turned up https://people.math.gatech.edu/~heil/6338/summer08/section9f.pdf, where Heil says on page 363 that weakly compact does not imply weakly metrizable, even in a separable space. But his proposed counterexample is the closed unit ball of $\ell^1$, which isn't actually weakly compact. And Daniel Fischer's proof looks good to me.) (Added: I had an email conversation with Professor Heil, who acknowledges that this is a mistake in the notes.)