From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ (the Banach space of the trace class operators equipped with its natural norm topology induced by the notion of trace) is the topological dual space of $B_\infty(H)$ (the Banach space of compact operators with the natual norm operator topology).
The duality is represented in terms of the trace operation as $B_1(H)$ is a (both sides *) ideal, as the (left-continuous) pairing
$$\langle A, B \rangle := tr(AB)\:,\:\quad A\in B_1(H)\:, \quad B \in B_\infty(H)$$
The Banach-Alaouglu theorem proves that every closed ball in $B_1(H)$ is therefore compact with respect to the weak * topology induced by the topological duality considered above. In particular, according to this weak* topology, $$B_1(H) \ni \rho_n \to \rho \in B_1(H)\quad iff \quad tr((\rho_n-\rho)A)\to 0 \quad \forall A\in B_1(H)$$ To conclude the proof it would be sufficient to prove that $S(H)$ is closed with respect to that topology, since closed subsets of compact sets are compact as well.
But it is not the case as proved here!
On the other hand, since weak * topology is Hausdroff, compact sets are closed. This proves that the assertion, referred to the said weak * topology, is false as $S(H)$ is not closed with respect to that weak * topology.
Maybe the assertion is true using another weak (natural?) topology.