If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$
Previously in the same theorem actually, they proved the following:
If $C$ is a choice rule satisfying the weak axiom, then $\succsim^*$ is rational, and for all $B \in \mathcal{P}(X), C(B) = C^*(B, \succsim^*).$
Note: $C^*(B) = C^*(B, \succsim) = \{ x \in B: \forall y \in B, x \succsim y \}$, and the weak axiom is referring to the weak axiom of revealed preference which a choice rule satisfies if $[x \succsim^* y ] \Rightarrow \neg (\exists B)[y \succ_B x]$
My try of the proof:
First suppose $B \mapsto C^*(B, \succsim)$ does not satisfy the weak axiom, then $\succsim^*$ is not rational by the earlier part of the theorem, so if we can prove $\succsim = \succsim^*$ we have a contradiction which will prove the statement.
As $\succsim$ is rational, then there is a utility function $u: X \to \mathbb{R}$ that represents $\succsim$, then $\{x \in B: \forall y \in B, u(x) \geq u(y) \} = C^*(B, \succsim)$ but then $\exists B$ s.t. $x,y \in B$ and since $B \mapsto C^*(B, \succsim)$ then $u(x) \in C^*(B, \succsim)$ so $x \in C(B)$ hence $\succsim = \succsim^*$