There's a typo when you say f(c)/1-f(1-c) > A/B, the correct would be f(c)/1-f(1-c) > B/A by the enunciate.
First you assumed you had A > B, which tell us for any values of A and B that the fraction B/A is always less than 1. So, f(c)/1-f(1-c) > B/A implies f(c)/1-f(1-c) < 1.
You showed also that f(c)/1-f(1-c) < 1 for any c in (0,1) by the convex propriety. Since you have f(0)=0 and f(1)=1 and all the other proprieties of this kind of function, you can say that f(c)/1-f(1-c) can be any number between the interval (0,1). As B/A is also a number between (0,1), by saying that exists a c* which makes f(c*)/1-f(1-c*) < B/A you are proving that there exists at least one c in (0,1) which makes the first inequality false (because you started with the premise f(c)/1-f(1-c) > B/A) and this completes the proof, because it is the contrapositive of the first affirmative (not for all c's implies that exists a c which does not hold the inequality).
I hope it helps.