uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$ If we assume $AC$ we can construct an uncountable well-ordered chain of subsets of $\mathbb{N}$ by well-ordering the reals and then using the same Dedekind's cut construction as in this question.

edit: as pointed out in the comments and Asaf's answer the construction above doesn't work

What if we don't assume $AC$? Can we show the existence of an uncountable, well-ordered chain of subsets of $\mathbb{N}$ in $ZF$ alone?
You're confusing well-ordered with well-orderable.

There is no uncountable well-ordered chain in $\mathcal P(\Bbb N)$, because if there was such $\{A_\alpha\mid\alpha&lt;\omega_1\}$, map $\alpha$ to the least $n$ in $A_{\alpha+1}\setminus A_\alpha$, and you got yourself an injection from $\omega_1$ into $\omega$. Something which you cannot get with or without choice.

But it is also consistent without the axiom of choice that there is no uncountable well-orderable chain in $\mathcal P(\Bbb N)$, as it is consistent that there is no injection from $\omega_1$ into $\mathcal P(\Bbb N)$ which means every well-orderable subset is countable. So we cannot prove in $\sf ZF$ the existence of an uncountable well-orderable subset of $\mathcal P(\Bbb N)$, chain or otherwise.
There is none since the set of integers is countable. The dedekind cut produces an uncountable chain which isn't well ordered.