Let $\alpha, \beta$ be ordinals and $n$ be a natural number. How can one prove the following $$ \forall \alpha (\alpha \approx n \implies \alpha = n)$$ using the fact $$ \vert \alpha \vert \le \beta \le \alpha \implies \vert \alpha \vert = \vert \beta \vert$$
and not using the fact that $n = \vert n \vert$.
Suppose $\alpha \approx n$, then $\vert \alpha \vert = \vert n\vert $.
If $n<\alpha$, then $\vert \alpha \vert = \vert n\vert \leq n <n+1 \leq \alpha$, so, by the Lemma, $\vert n+1 \vert = \vert \alpha \vert = \vert n \vert$, a contradiction of $n\not\approx n+1$.
If $\alpha <n$, then $\vert n \vert = \vert \alpha\vert \leq \alpha <\alpha+1 \leq n$, so $\vert \alpha +1\vert= \vert n\vert =\vert \alpha\vert$, but there is a bijection between $\alpha +1$ and $n+1$(because there is a bijection between $\alpha$ and $n$), then $\vert n+1\vert =\vert \alpha +1 \vert =\vert n\vert$, again, a contradiction of $n\not\approx n+1$.
Therefore $\alpha=n$.