Assuming countable choice, every infinite sigma-algebra contains a copy of the power set of the natural numbers

For each $m\ge 1$, let $I_m$ be the nonempty set of injections $m\to \mathcal{A}$. By [L1] choose $f_m\in I_m$. By [L6], the union of the finite ranges $f_m[m]$ is at most countable. It is infinite because it has subsets of every finite size, so [L5] makes it countably infinite; a bijective listing is therefore an injective sequence in $\mathcal{A}$.

By [L2], fix pairwise disjoint nonempty $D_n\in \mathcal{A}$. For $S\subseteq \mathbb{N}$, define $\mathrm{\Phi }(S):={\bigcup }_{n\in S}{D}_n$, which lies in $\mathcal{A}$ by [L3].

If $S\mathrm{\ne }T$, the least index in $S\mathrm{△}T$ belongs to exactly one of them, and its nonempty $D_n$ is contained in exactly one of $\mathrm{\Phi }(S),\mathrm{\Phi }(T)$ by disjointness. Thus $\mathrm{\Phi }$ is injective. If $\mathcal{A}$ were at most countable, [L5] would give an injection $j:\mathcal{A}\to \mathbb{N}$; the map that sends $j(\mathrm{\Phi }(S))$ to $S$ and every natural number outside $j[\mathrm{\Phi }[\mathcal{P}(\mathbb{N})]]$ to $\mathrm{\varnothing }$ would then be a surjection $\mathbb{N}\to \mathcal{P}(\mathbb{N})$, contrary to [L4]. Hence $\mathcal{A}$ is uncountable.