There is a problem in step 9, but the problem is not related to introducing a third person. The existence of a third person makes step 9 correct.
The problem is that we establish the base cases S(0) and S(1), and then we establish the inductive step S(n) -> S(n+1) subject to the restriction that n > 1. But this proof never demonstrates that S(1) implies S(2). The proof is completely correct that S(2) implies S(3), S(3) implies S(4), and so on.
When step 9 says "Let R be someone else in G other than P or Q", the existence of a person in G other than P or Q has not been demonstrated. G is an arbitrary group of k+1 people, and the highest k for which our statement is known to be true is 1. Thus, the minimum size of G is 2, and R may not exist.
The problem is that we establish the base cases S(0) and S(1), and then we establish the inductive step S(n) -> S(n+1) subject to the restriction that n > 1. But this proof never demonstrates that S(1) implies S(2). The proof is completely correct that S(2) implies S(3), S(3) implies S(4), and so on.
When step 9 says "Let R be someone else in G other than P or Q", the existence of a person in G other than P or Q has not been demonstrated. G is an arbitrary group of k+1 people, and the highest k for which our statement is known to be true is 1. Thus, the minimum size of G is 2, and R may not exist.