I am studying with Jech's book, and the above statement is stated without any proof. I want to know whether my proof is right and also I want to know if I am missing some points that makes this trivial;the proof is redundant.
Let $f\colon \lvert \alpha \rvert \to \alpha$ be the bijection. Let's define a property on $\mathrm{Ord}$ $$P(\beta) := \text{there exists a sequence } \langle a_\nu : \nu < \beta \rangle \text{ such that } a_\nu = \inf\{\gamma: f(\gamma) > f(a_\xi) \text{ for all } \xi < \nu\} \text{ for all } \nu < \beta$$ If $P(\beta)$, the corresponding sequence is unique by well-founded induction. The sequence $\langle a_\nu : \nu < \beta \rangle$ is trivially increasing thus an injection, so $\beta \le \lvert \alpha \rvert$.
Then $\theta = \{\beta \in \mathrm{Ord} : P(\beta)\}$ is a set, and also an ordinal. For all $\beta < \theta$, let $h_\beta\colon \beta \to \lvert \alpha \rvert$ be the corresponding sequence. Then $h = \bigcup_{\beta < \theta}h_\beta$ is a function, also proved by well-founded induction. Then $P(\operatorname{dom} h)$ so $\operatorname{dom} h + 1 = \theta$.
If $\sup \operatorname{ran} f \circ h < \alpha$, let $\delta = \inf \{ \gamma : f(\gamma) > \sup \operatorname{ran} f \circ h \}$. Then $h \cup \{(\operatorname{dom} h, \delta)\}$ satisfies the condition so $\theta = \operatorname{dom} h + 1 \in \theta$, which is a contradiction. Then $f \circ h\colon \operatorname{dom} h \to \alpha$ is cofinal in $\alpha$, and automatically $\operatorname{dom} h$ is a limit ordinal. Finally $\operatorname{cf} \alpha \le \operatorname{dom} h \le \lvert \alpha \rvert$.
Assume towards a contradiction that we have an ordinal $\alpha$ where $\textrm{cf }\alpha>\alpha$. $\alpha$ itself is a cofinal subset of $\alpha$ that is unbounded in $\alpha$, and it has order type $\alpha$ under $<$. But this contradicts $\textrm{cf }\alpha>\alpha$.