The assertion is **false**, if I understand well it.

First of all some general information on the notion of weak * topology.

One has a (complex) normed space $B$ and its topological dual $B'$ which is, by definition the vector space of continuous linear maps $f: B \to \mathbb{C}$.

This space has a natural structure of normed space as well with the norm
$$||f||' := \sup\{|f(x)| \:|\: x \in B \:, ||x||=1\}\:.$$ 
However we can equip $B'$ with another seminormed topology induced by $B$ called **weak * topology**, it is characterized by the fact that 

$$B' \ni f_n \to f \in B'\quad iff \quad f_n(x)-f(x)\to 0 \quad \forall x\in B\:.$$
 
The **Banach-Alaoglu theorem** proves that every norm-closed ball in $B'$
$$\{f \in B' \:|\: ||f|| \leq r\}$$
is compact with respect to the weak * topology provided $B$ is complete (i.e. is a Banach space).

Sometimes it happens that two normed spaces $B_1$ and $B_2$ are in topological duality, that is one is (normed-space) isomorphic to the topological dual of the other. If $B_2$ is isomorphic through $F$  to $B_1' = F(B_2)$, the action of the elements of $B_2$ o the elements of $B_1$ is indicated by means of a pairing:
$$\langle b_1, b_2\rangle := (F(b_2))(b_1)\:.$$


 

Let us pass to the discussed assertion, that the space of the mixed states $S(H)$ is compact with respect to some weak * topology on that. The space $S(H)$ is not linear and it is not a closed ball of
the normed (Banach) space of the trace class operators $B_1(H)$ equipped with its natural norm topology induced by the notion of trace.
 Therefore, one should try to prove that  $S(H)$ is a compact subset of $B_1(H)$.

From the sketch of proof, it seems that the authors use the fact that $B_1(H)$ is the topological dual space of some other space. All that in order to apply the Banach-Alaoglu theorem as written above.(*)

As is well known, $B_1(H)$ is in fact (isometrically isomorphic to) 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-Alaoglu 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][1]!

On the other hand, *since weak * topology is Hausdorff*, 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. 

---------------------------
($*$) Another possibility, not followed by the authors, is to see $B_1(H) \subset B(H)$ and viewing $B(H)$ as the dual of $B_1(H)$, thus equipping this latter with the weak* topology of $B(H)$ also known as ultraweak topology.

Even another way is to see $B_1(H)$ as subset of $B_2(H)$ the Hilbert-space space of Hilbert Schmidt operators and to use the weak * topology  on $B_2(H)$ induced by the Riesz anti isomorphism. 

  [1]: https://math.stackexchange.com/questions/4783685/compactness-of-subset-of-trace-class-operators-on-a-hilbert-space