If $\succsim$ is rational, then $B \mapsto C^*(B, \succsim)$ satisfies the weak axiom, and $\succsim=\succsim^*$
Here is an attempt at a direct proof:

Let all the assumptions hold, and further assume that $x \succcurlyeq^* y$. Then for some $B \supseteq \{x,y\}$, $C^*(B) \supseteq \{x\}$ (by definition of $\succcurlyeq^*$), which further implies $x \succcurlyeq z$ for all $z \in B$, and therefore $x \succcurlyeq y$ (by definition of $C^*$). (This shows that $\succcurlyeq^* \subseteq \succcurlyeq$; the other direction follows from looking at doubleton sets $B = \{x,y\}$).

What is left to show is that $C^*$ satisfies the weak axiom. So take any $B' \supseteq \{x,y\}$, such that $y \in C^*(B')$. If no such set exists, then $y$ is not revealed preferred to $x$ and we are done. If such a set exists then by definition $y \succcurlyeq z$ for all $z \in B'$. We already know $x \succcurlyeq y$ so by transativity $x \succcurlyeq z$ for all $z \in B'$ and hence $x \in C^*(B')$. Again,  $y$ is not revealed preferred to $x$ and we are done.

Notice, while we used transitivity, we did not use completeness. This hints at the fact that the statement can actually be generalized to pre-orders (reflexive and transitive), by allowing $C^*$ to be empty. This point is somewhat lost by appealing to the second result.