The question is stated as:
Prove the Principle of Complete Induction
$$(\forall B)([B \subseteq P \land (\forall x)((\forall y)(y<x \Rightarrow y \in B)\Rightarrow x \in B)] \Rightarrow P=B)$$
(Hint: Assume $P \neq B$. Then, $P-B \neq \emptyset$. Apply the Least Number Principle. Notice that, when $x=1$, $(\forall y)(y<x \Rightarrow y \in B)$ is trivially true, since $y<1$ is always false.)
The set $P$ here is a set from a Peano System $(P,S,1)$.
Here is my atempt:
Let $B$ be a some subset of $P$ where $(\forall x)((\forall y)(y<x \Rightarrow y \in B)\Rightarrow x \in B)$ holds true, the if $x=1$ we get $(\forall y)(y<1 \Rightarrow y \in B)$ which is trivially true, then $1 \in B$.
Now assume $x \in B$ thus, $(\forall y)(y < x \Rightarrow y \in B)$. If we take $S(x)$ we have that $(\forall y)(y < x < S(x))$, but as we have $x \in B$ by assumption we have also $(\forall y)(y < S(x) \Rightarrow y \in B)$ which implies $S(x) \in B$, and therefore $x \in B \Rightarrow S(x) \in B$, then by mathematical induction $P=B$.
Edit 1: Trying to use the author hint
Let $B$ be a subset of $P$ where, $(\forall x)((\forall y)(y<x \Rightarrow y \in B)\Rightarrow x \in B)$ holds true.
Assume $B \neq P$ and let $A=P-B$, so $A \neq \emptyset$, as $A \subseteq P \land A \neq \emptyset$ then $A$ have a least element that is $(\exists z)(z \in A \land (\forall u )(u \in A \Rightarrow z \leq u))$
We have as a Theorem in Peano System that $(\forall x)(x=1 \lor (\exists p)(x=S(p)))$
Thus taking $z$ as the least element of $A$ we have that $z=1 \lor (\exists p)(z = S(p))$, if we suppose $z=1$, then $1 \in A$, so $1 \notin B$, thus $(\exists y)(y<1 \land y \notin B)$ which is clearly false, thuse we have that $z \neq 1$.
But if $z=S(p)$ for some $p$ in $P$, we have that $S(p) \in A$ then $S(p) \notin B$ and from this $(\exists y)(y<S(p) \land y \notin B)$, but if $y \notin B$ we have that $y \in A$. But if $(y<S(p) \land y \in A)$ then $S(p)$ cant be the least element of $A$ which is a contradiction.
Edit 1 end
I think I have missed the point in this exercise, because I cant manage to use the Least Number Principle at all and that was the hint from the author, my proof is valid? and how can I use ths Least Number Principle in this case?
Best Answer
Here is the argument that the OP is looking for (we've eliminated the quantifiers and formal presentation to focus on the main proof 'flow').
The set $B$ satisfies the (complete/strong) induction hypotheses.
Assume $B \subsetneq P$ and let
$\tag 1 m = \text{min}(P \setminus B)$
Then
$\tag 2 x \lt m \Rightarrow x \in B$
is true (otherwise $m$ wouldn't be the minimum number).
By the inductive hypyotheis $m\in B$, but that (also) contradicts $\text{(1)}$.
So assuming $B \subsetneq P$ leads to a contradiction, and so $B = P$.