Many books indicate yes to this question. However, I found the only lemma they claim to use AC is the following statement:
If $\{A_i\}_{i\in I}$ is a family of sets, then $|\bigcup_{i\in I}A_i|\leq|I|\sup_{i\in I}|A_i|$.
But if we replace this statement by the following version, then this does not require AC anymore.
If $\{\beta_\alpha\}_{\alpha<\gamma}$ is a sequence of ordinals, then $|\sup_{\alpha<\gamma}\beta_\alpha|\leq|\gamma|\sup_{\alpha<\gamma}|\beta_\alpha|$.
Proof: We map $\delta$ to the least $\alpha$ such that $\delta<\beta_\alpha$ together with the image of $\delta$ under the map $f:\beta_\alpha\to|\beta_\alpha|\to\sup_{\alpha<\gamma}|\beta_\alpha|$, i.e., $\delta\mapsto(\alpha,f(\delta))$.
This statement is used to prove:
For infinite $\kappa$, $\mathrm{cf}(\kappa)$ is the least cardinal $\lambda$ such that $\kappa=\sum_{\beta<\lambda}\kappa_\beta$, where $\kappa_\beta<\kappa$.
Proof:
($\leq$) Let $\kappa=\sum_{\alpha<\lambda}\kappa_\alpha$, where $\kappa_\alpha<\kappa$. We may assume $\kappa_\alpha\geq 1$ for all $\alpha<\lambda$. Then either $\lambda$ or one of $\kappa_\alpha$ is infinite. So $\kappa=\lambda\sup_{\alpha<\lambda}\kappa_\alpha$. If $\kappa=\lambda$, then $\mathrm{cf}(\kappa)\leq\lambda$. If $\kappa=\sup_{\alpha<\lambda}\kappa_\alpha$, then we have a cofinal sequence, so $\mathrm{cf}(\kappa)\leq\lambda$ in this case as well.
($\geq$) Let $f:\mathrm{cf}(\kappa)\to\kappa$ be a cofinal sequence and write $\lambda=\mathrm{cf}(\kappa)$. We may assume $f$ is nonzero. Then we see $\kappa=|\kappa|=|\sup_{\alpha<\lambda}f(\alpha)|\leq\lambda\sup_{\alpha<\lambda}|f(\alpha)|\leq\kappa$. And since $\lambda\sup_{\alpha<\lambda}|f(\alpha)|=\sum_{\alpha<\lambda}|f(\alpha)|$, we got the equality $\kappa=\sum_{\alpha<\lambda}|f(\alpha)|$.
Then the proof goes:
If $\omega_{\alpha+1}$ were singular, then $\lambda=\mathrm{cf}(\omega_{\alpha+1})\leq\omega_\alpha$. Write $\omega_{\alpha+1}=\sum_{\beta<\lambda}\kappa_\beta$, where $\kappa_\beta\leq\omega_\alpha$. Hence, $\omega_{\alpha+1}=\sum_{\beta<\lambda}\kappa_\beta\leq\sum_{\omega_\alpha}\omega_\alpha=\omega_\alpha^2=\omega_\alpha$, contradiction.
Maybe I made a mistake somewhere, I don't know. Could anyone identify where is AC hiding in this proof? Thanks!
Disclaimer: all cardinals are well-orderable ordinals.
You have a hidden assumption in the proof of your first proposition. If $\beta_{x}\leq d<d'<\beta_{\alpha}$ whenever $x<\alpha ,$ then in order that the map that sends $d$ to $(\alpha,f_{\alpha}(d))$ be 1-to-1, you need each$f_{\alpha}:\beta_{\alpha}\to |\beta_{\alpha}|$ to be 1-to-1. Now for each $\alpha$, such $f_{\alpha}$ exists but it is generally not unique, so we cannot apply the Replacement Axiom to assert the existence of the sequence $(f_{\alpha})_{\alpha < \beta}.$ This is where AC is hiding.
Even when trying to prove in ZF that $\omega_1$ is regular, we hit this difficulty when we try to show in ZF that a countable union of countable ordinals is countable.
I dk whether there has been any progress on the open problem of whether it can be shown in ZF that a regular uncountable cardinal exists.