Proper definition of the natural numbers.
Here's one taken (from memory) from Halmos's Naive Set Theory.
Definition. Given a set $x$, we define $x^+$, the successor of $x$, to be the set $x\cup\{x\}$.
If $x$ is a set, $x^+$ is a set: $\{x\}$ is an element of the power set of $X$, so $x^+$ is a subset of $x\cup\mathcal{P}(x)$, hence a set (by the Axiom of Powers, Axiom of Unions, and Axiom of Separation).
Definition. A set $S$ is said to be inductive if and only if $\emptyset\in S$ and for every $x$, if $x\in S$ then $x^+\in S$.
Axiom of Infinity. There exists at least one inductive set.
Now, let $S$ be any inductive set. Let
$$\mathbb{N}_S = \bigcap\{A\subseteq S\mid A\text{ is inductive}\}.$$
This is a set, since the family is a subsets of $\mathcal{P}(S)$, and so its intersection is a set.
Lemma. The intersection of any nonempty collection of inductive sets is inductive.
Proof. Suppose $\{S_i\}$ is a nonempty family of inductive sets, and let $S=\cap S_i$ Then $\emptyset\in S_i$ for each $i$, hence $\emptyset\in S$; and if $x\in S$, then $x\in S_i$, for each $i$; hence (since each $S_i$ is inductive), $x^+\in S_i$ for each $i$, and thus $x^+\in S$. Thus, $S$ is inductive. $\Box$
Corollary. If $S$ is inductive, then $\mathbb{N}_S$ is inductive. Moreover, if $B$ is any inductive subset of $S$, then $\mathbb{N}_S\subseteq B$.
Theorem. If $S$ and $T$ are two inductive sets, then $\mathbb{N}_S = \mathbb{N}_T$.
Proof. Consider $\mathbb{N}_T\cap S$; this is an inductive subset of $S$, hence $\mathbb{N}_S\subseteq \mathbb{N}_T\cap S\subseteq \mathbb{N}_T$. Symmetrically, since $\mathbb{N}_S\cap T$ is inductive, then $\mathbb{N}_T\subseteq \mathbb{N}_S\cap T\subseteq\mathbb{N}_S$. Thus, $\mathbb{N}_S=\mathbb{N}_T$. $\Box$
Definition. $\mathbb{N}$ is the set $\mathbb{N}_S$, where $S$ is any inductive set.
This set is well defined, since by the theorem above it does not matter what inductive set we start with, we always get the same set $\mathbb{N}$; and there is at least one inductive set by the Axiom of Infinity.
Theorem. If $S$ is any inductive set, then $\mathbb{N}\subseteq S$; that is, $\mathbb{N}$ is the "smallest" inductive set, in the sense of set inclusion.
Proof. If $S$ is any inductive set, then $\mathbb{N}=\mathbb{N}_S\subseteq S$. $\Box$
Now let $0=\emptyset$, and let $s(x) = x^+$ for every $x$.
Theorem. (The Peano Axioms) $\mathbb{N}$ satisfies the following:
- $0\in \mathbb{N}$.
- If $n\in \mathbb{N}$, then $s(n)\in\mathbb{N}$.
- For all $n\in\mathbb{N}$, $s(n)\neq 0$.
- If $s(n)=s(m)$, then $n=m$.
- If $S\subseteq \mathbb{N}$ is such that $0\in S$ and if $n\in S$ then $s(n)\in S$, then $S=\mathbb{N}$.
Proof. Since $\mathbb{N}$ is inductive, 1 and 2 are satisfied. 3 is satisfied because by definition, $n\in s(n)$, hence $s(n)\neq\emptyset=0$. 5 is satisfied because the given conditions imply that $S$ is inductive, and therefore $\mathbb{N}\subseteq S$. Since we already have $S\subseteq \mathbb{N}$, then $S=\mathbb{N}$ holds.
The only difficulty is property 4. This requires some auxiliary results:
Lemma 1. If $n\in\mathbb{N}$, and $m\in n$, then $m\subseteq n$.
Proof. Let $S=\{n\in\mathbb{N}\mid \text{for all }m\in n, m\subseteq n\}$.
Then $0\in S$, since $0=\emptyset$ satisfies the property by vacuity. Assume that $k\in S$, and let $m\in s(k) = k\cup\{k\}$. If $m\in k$, then since $k\in S$ we have $m\subseteq k\subseteq s(k)$. If $m\notin k$, then $m\in\{k\}$, hence $m=k$, and $m=k\subseteq k\cup\{k\}=s(k)$. Thus, $s(k)\in S$.
We conclude that $S$ is an inductive subset of $\mathbb{N}$, and therefore that $S=\mathbb{N}$. $\Box$
Lemma 2. If $n\in\mathbb{N}$ and $m\in n$, then $n\not\subseteq m$.
Proof. Let $S=\{n\in\mathbb{N}\mid\text{for all }m\in n, n\not\subseteq m\}$. Then $0\in S$, since it satisfies the condition by vacuity.
Assume that $k\in S$. If $m\in s(k)=k\cup\{k\}$, then either $m\in k$, in which case $k\not\subseteq m$, hence $s(k)\not\subseteq m$ (since $k\subseteq s(k)$); or else $m=k$. But since $k\in S$, then $k\notin k$ (for $k\subseteq k$, so $k\in k$ would contradict that $k\in S$). Since $k\in s(k)$, then $s(k)\not\subseteq k=m$. Thus, if $k\in S$ then $s(k)\in S$. Hence $S$ is an inductive subset of $\mathbb{N}$, so $S=\mathbb{N}$. $\Box$
We are now ready to prove property 4. Assume that $s(n)=s(m)$. Since $m\in s(m)=s(n) = n\cup \{n\}$, then either $m=n$, or $m\in n$. If $m=n$, we are done. So assume that $m\in n$. Since $n\in s(n)=s(m)$, then $n\in m\cup \{m\}$; since $n\neq m$, then $n\in m$. Thus, $n\in m$, hence $n\subseteq m$; but $m\in n$, contradicting Lemma 2. This contradiction arises from assuming that $m\in n$. Therefore, $m\notin n$, so $m=n$, hence $\mathbb{N}$ satisfies the fourth Peano axiom. $\Box$
Proof that $\lt$ is a total order on $\mathbb{N}$.
(I'm more used to the definition $n\lt m\Longleftrightarrow n\in m$, but I'll use your definition with proper inclusion)
We want to prove that if $n,m\in\mathbb{N}$, then either $n=m$, $n\subset m$, or $m\subset n$.
Lemma. Let $n\in \mathbb{N}$. Then either $n=0$, or there exists $k\in\mathbb{N}$ such that $n=s(k)$.
Proof. Let $S=\{n\in\mathbb{N}\mid n=0\text{ or }n=s(k)\text{ for some }k\in\mathbb{N}\}$.
Then $0\in S$; if $k\in S$, $s(k)\in S$ (since $s(k)$ is a successor). Thus, $S$ is inductive, hence $S=\mathbb{N}$. $\Box$
Lemma. Let $k,n\in\mathbb{N}$. If $k\subset n$, then $s(k)\subseteq n$.
Proof. Let $S=\{n\in\mathbb{N}\mid \text{if }k\subset n\text{ then }s(k)\subseteq n\}$.
Then $0\in S$ by vacuity. If $n\in S$, let $k\subset s(n)=n\cup\{n\}$.
If $k\subset n$, then $s(k)\subseteq n\subset s(n)$. If $k=n$, then $s(k)=s(n)$. The only other possibility is that $n\in k$. But since $k$ is a natural number, then $n\in k$ implies $n\subseteq k$. But $n\subseteq k\subset n\cup\{n\}$ is impossible. Hence, $k\subset s(n)$ simplies $s(k)\subseteq s(n)$. This proves $S$ is inductive, hence $S=\mathbb{N}$.
Finally, fix $m$, and let $S=\{n\in\mathbb{N}\mid m\subseteq n\text{ or }n\subset m\}$.
Then $0\in S$: if $m=\emptyset$, then $m\subseteq 0$; if $m\neq\emptyset$, then $\emptyset\subset m$.
Assume that $k\in S$. If $m\subseteq k$, then $m\subseteq s(k)$. If $k\subset m$, then by the Lemma $s(k)\subseteq m$. If $s(k)=m$, then $s(k)\in S$. If $s(k)\subset m$, then $s(k)\in S$. Either way, $S$ is inductive, hence $S=\mathbb{N}$. Therefore, $\lt$ is a trichotomic relation on $S$.
Proof of the Well Ordering Principle
Lemma. If $n\in\mathbb{N}$, then there does not exist $k\in\mathbb{N}$ such that $n\lt k\lt s(n)$.
Proof. It is impossible to have $n\subset k\subset n\cup\{n\}$. $\Box$
Well Ordering Principle. If $A\subseteq\mathbb{N}$ and $A\neq\emptyset$, then $A$ has a smallest element.
Proof. Let $A\neq\emptyset$, $A\subseteq \mathbb{N}$. Let $S=\mathbb{N}-A$.
Let $S'= \{n\in\mathbb{N}\mid n\in S\text{ and for all }k,\text{ if }k\lt n\text{ then }k\in S\}$. Since $S'\subseteq S\neq\mathbb{N}$, then $S'$ is not inductive. Therefore, either $0\notin S'$, or else there exists $k\in S'$ such that $s(k)\notin S'$.
If $0\notin S'$, then since for all $k\lt 0$, $k\in S$, it follows that $0\notin S$. Therefore, $0\in A$. Since $0$ is the smallest element of $\mathbb{N}$, it is also the smallest element of $A$ and we are done.
Assume then that there exist $k\in \mathbb{N}$ such that $k\in S'$ but $s(k)\notin S'$. Then $k\in S$ and for all $n\in\mathbb{N}$, if $n\lt k$ then $n\in S$. That means that for all $n\in\mathbb{N}$, if $n\lt s(k)$, then $n\in S$ (for $n\lt s(k)$ implies $n\leq k$ by the Lemma, and so either $n=k\in S$ or $n\lt k$ hence $n\in S$ since $k\in S'$). That means that because $s(k)\notin S'$, it must be because $s(k)\notin S$. Thus, $s(k)\in A$. However, as we have already seen, if $n\lt s(k)$, then $n\in S$, hence $n\lt s(k)\Rightarrow n\notin A$. Thus, $n\in A\Rightarrow s(k)\leq n$, which proves that $s(k)$ is the smallest element of $A$. $\Box$
