Page 31 of Jech's "Set Theory" (3rd edition), for the proof of theorem 3.5:
[...] let $\alpha$ be the least ordinal such that $\Gamma(\omega_\alpha\times\omega_\alpha ) \neq \omega_\alpha$.
Let $\beta, \gamma < \omega_\alpha$ be such that $\Gamma(\beta,\gamma) = \omega_\alpha$.
Pick $\delta < \omega_\alpha$ such that $\delta > \beta$ and $\delta > \gamma$
where $\Gamma$ is a one-to-one mapping of $Ord^2$ onto $Ord$ defined by letting $\Gamma(\alpha,\beta)$ be the order type of the set $\{(\xi, \eta) : (\xi, \eta) < (\alpha, \beta)\}$ in the canonical well-ordering of $Ord^2$.
I don't see why we can necessarily find $\beta$ and $\gamma$ both strictly inferior to $\omega_\alpha$, such that $\Gamma(\beta,\gamma)=\omega_\alpha$.
Why couldn't one of them, say $\beta$, be equal to $\omega_\alpha$, and only the other one strictly inferior to it?
A straightforward induction argument shows that $\Gamma(\beta\times\beta)$ is an ordinal $\ge \beta$ for all $\beta,$ so if $\omega_\alpha\ne \Gamma(\omega_\alpha\times\omega_\alpha)$, this means $\omega_\alpha\in \Gamma(\omega_\alpha\times\omega_\alpha),$ which means precisely that there are $\beta,\gamma < \omega_\alpha$ with $\Gamma(\beta,\gamma)=\omega_\alpha.$