Let $(P,Sc,1)$ a Peano's system, then $P=\{1\}\cup Sc\{P\}$
They use the third Peano's axiom, in which if $A\subseteq P, 1\in A$ and $Sc(a)\subseteq A\Rightarrow A=P$. But their proof says in the begining: It's enough to prove that $Sc(A)\subseteq A$
How do they know that $A\subseteq P$?
By definition, $P$ is closed under sucessors, i.e., $\operatorname{Sc}(P)\subseteq P$. Also, by definition $1\in P$, so $\{1\}\subseteq P$.
So if we define $A=\{1\}\cup\operatorname{Sc}(P)$, it is the union of two subsets of $P$, hence a subset of $P$ itself. The proof goes on to showing that $1\in A$ and $\operatorname{Sc}(A)\subseteq A$, from which the third Peano axiom (induction) implies that $P=A=\{1\}\cup\operatorname{Sc}(P)$.