First note that by definition of $d(x)$ we have $f_0(x)/f_A(x)\ge c$ when $d(x)=0$ and $f_0(x)/f_A(x)\lt c$ when $d(x)=1$. These inequalities are equivalent to $c f_A(x)-f_0(x)\le 0$ and $c f_A(x)-f_0(x)\gt 0$ respectively.
Now consider the case where $d(x)=0$. Then the right hand side of the inequality to be proven is zero and the left hand side is non-positive. The latter follows from the fact that $d^*(x)\ge 0$ and also because we have seen that $c f_A(x)-f_0(x)\le 0$ when $d(x)=0$. So the inequality holds when $d(x)=0$.
Next consider the case where $d(x)=1$ and $d*(x)=0$. In this case, the right hand side of the inequality to be proven is $c f_A(x)-f_0(x)$, which must be positive since $d(x)=1$. Since the left hand side is zero in this case, again the inequality holds.
The last possibility is the case where both $d(x)=1$ and $d^*(x)=1$. In this case both sides of the inequality to be proven are equal to $c f_A(x)-f_0(x)$, and so again the inequality holds.
QED
BTW, you are correct about the typo.