How can we show $n \subset m \rightarrow n \in m \lor n=m$? I want to prove that for any natural numbers $n,m$ it holds that:
$$n \subset m \leftrightarrow n \in m \lor n=m$$
$"\Leftarrow"$: If $n \in m \lor n=m$, using the sentence: For any natural numbers $m,n$ it holds that $n \in m \rightarrow n \subset m$, we conclude that $n \subset m$.
$"\Rightarrow"$:Could we  justify it like that?
$$n \subset m \rightarrow n=m \lor n \subsetneq m$$
When $n=m$, the desired property is satisfied.
When $n \subsetneq m$ How could we show that $n=m \lor n \subsetneq m$ ? Could we show it using induction? If so, how could we do this?
 A: Proposition 1: $\{n\} \in m \to n \in m$
Proof: For $m=\emptyset$ we are okay since there is no $\{n\}\in m$.
Now let $\{n\}\in m \to n\in m$. We must show that $\{n\}\in S(m) \to n\in S(m)$.
$$S(m) = m\cup \{m\}$$
Thus $\{n\} \in S(m) \to \{n\}\in m \vee \{n\} = \{m\}$. In the first case we have by condition $n\in m$ and since $m\subset S(m)$, $n\in S(m)$. The second case $\{n\} = \{m\} \to n=m \to n\in S(m)$, thus $n\in S(m)$.
$\qquad\Box$
Proposition 2: $n\subset m \to n \in m \vee n = m$
Proof: If $n=m$ then clearly $n\subset m$, now assume $n\subset m$ and $n\ne m$, i.e. $n\subsetneq m$. Now if $n\subsetneq m$ then $\{n\}\in m$ as you said. Finally by Proposition 1, $n\in m$.
Alternative last step without Proposition 1: Since $n\subsetneq m$, $n \cup \{n\} = S(n) \subset m$ so especially $n\in S(n)\subset m$, i.e. $n\in m$ as claimed.
A: Lemma 1: Every natural number is a transitive set.
Proof: Induction. $0=\varnothing$, so it's trivially transitive. Assume $n$ is transitive. Then $S(n)=n\cup\{n\}$ is transitive: let $k\in m\in S(n)$. Then either $m\in n$ or $m=n$. If $m\in n$, then since $n$ is transitive, $k\in n\subsetneq S(n)$, so $k\in S(n)$. Otherwise, $m=n$, in which case $k\in n$ directly, and thus $k\in S(n)$. By induction, every natural number is transitive.
Note: Being transitive is equivalent to all elements also being subsets. 
Lemma 2: No natural number is an element of itself.
Proof: Induction. $0=\varnothing$ and $\varnothing\notin\varnothing$. Now assume $n\notin n$. $S(n)=n\cup\{n\}$. If $S(n)\in S(n)$, then either $S(n)=n$ or $S(n)\in n$. If $S(n)=n$, then $\{n\}\subseteq n\implies n\in n$, contradicting the inductive hypothesis. If $S(n)\in n$, then since $n\in S(n)$, and since $n$ is transitive, we have $n\in n$, again contradicting the inductive hypothesis. Thus, no natural number is an element of itself.
Lemma 3: $\omega$ is totally ordered by $\in$.
Proof: By Lemma 1, $\in$ gives a transitive relation on $\omega$. By Lemma 2, this relation is irreflexive. Thus, we need only show that for any $n, m\in\omega$, at least one of $m\in n, m=n, n\in m$ is true (transitivity and irreflexivity prevent more than one from being true). We fix $m$ and induct on $n$. Firstly, $0\in m$ or $0=m$. Next, if $n\in m$, then $S(n)\in m$ or $(n)=m$. If $n=m$, then $m\in S(n)$. Finally, if $m\in n$, then $m\in n\in S(n)\implies m\in S(n)$, since $S(n)$ is transitive. Thus, $m$ is comparable to any natural number. Since $m$ was arbitrary, any two natural numbers are comparable, so $\omega$ is totally ordered by $\in$. 
Theorem: If $n, m\in \omega$, then $n\subsetneq m\implies n\in m$.
Assume $n\subsetneq m$. We seek to show that $n\in m$. Since $\omega$ is totally ordered by elementhood, we have either $n\in m$ or $m\in n$. If $m\in n$, then since $n$ is transitive, all elements of $n$, including $m$, are subsets, so $n\subsetneq m\subseteq n\implies n\subsetneq n$, a contradiction.
