The Question
THEOREM 1.1 There is exactly one natural number which is not the successor of any natural number.
PROOF: Let $N_S$ be the range of $S$. By Axiom $A_2$, there exists a natural number $u$ which is not in $N_S$. Let $M=\{u\}\cup N_S$. Then $u\in M$ and $S(n)\in M$ for every $n\in M$, since $N_S\subset M$. By Axiom $A_3$, $M=N$. Hence $N=\{u\}\cup N_S$, and every natural number $n\ne u$ is in $N_S$, i.e., every natural number except $u$ is the successor of some natural number.
My Understanding
My trouble is that I can't understand the key point, that are the words "exactly one". If I add another natural number $u_2$ which is not in the range of $S$, it seems to me that the proof works fine too!!. I know that it must be something that I am losing. Can you help me?