An important results for natural numbers and their ordering by $<$ (that is, $\in$, $m < n$ means $m \in n$) is that for any natural numbers $m,n$ and $k$, we have $m < n \Longleftrightarrow m + 1 < n + 1$.
What preliminary results are needed? This is also a preliminary result, a lemma for the following result by induction: $\forall m,n,k \in \mathbb{N}, m < n \Longleftrightarrow m + k < n + k$.
Mathematical induction is assumed, of course, as is the fact that $m + 1 = m\cup\{m\}$. Also, it has already been proven that $\leq$ ($m \leq n$ stands for $m < n$ or $m = n$) is a well-ordering on $\mathbb{N}$. The addition is defined as follows: $\forall m,n \in \mathbb{N}$,
$m + 0 = m$,
$m + (n + 1) = (m + n) + 1$.
First of all, if $m+1<n+1$, then $m\cup\{m\}\in n\cup\{n\}$, therefore either $m\cup\{m\}=n$, in which case $m\in n$ and we are done, or $m\cup\{m\}\in n$, in which case $m\in n$ and we are done again.
Secondly, if $m\in n$, then either $m\cup\{m\}\in n$, in which case we are clearly done; or $m\cup\{m\}=n$, in which case $n\in n\cup\{n\}$ so $m\cup\{m\}\in n\cup\{n\}$.
What did we use? We used the fact that $n$ is transitive, so $x\in y\in n$ means $x\in n$. It is tempting that we directly appeal to the fact that $\in$ is well-founded, and therefore there are no $\in$-loops, but it doesn't matter. If they were are $\in$-loops in that definition it would just show that $<$ is not a linear ordering (or a partial ordering for that matter), the proof would still work.