For the first claim observe that none of the sets A-B can be empty:
Since B is a *minimal* blocking set of b(F), it cannot contain any 
member A of F as a proper subset, just because each member of F is 
a blocking set of b(F). Then argue as in the book. 
Note that this claim holds for *all* families F, not just for anti-chains.
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To prove the second claim, take any A in F.  We want to show A is in b(b(F)).  
Each element of b(F) intersects A, so A is a blocking set for b(F). 
Therefore A  contains (as a subset) some minimal blocking set B in b(b(F)).  
Since  b(b(F)) is a subset of F (by the first part of the proposition), 
the set B must belong to F.  Hence, A and its subset B are both in F.  
But F is an antichain, therefore A=B, so A in b(b(F)). Q.E.D.