A sigma-algebra with a listed infinite subfamily contains a disjoint sequence of nonempty members

Let $\mathcal{B}$ be the Boolean algebra generated by the sets $e(n)$. Finite Boolean expressions can be coded by natural numbers, so deleting repeated values in least-code order gives a listing of $\mathcal{B}$. It is infinite because it contains the distinct sets $e(n)$.

Call a nonempty $B\in \mathcal{B}$ an atom when it has no nonempty proper member in $\mathcal{B}$. If $\mathcal{B}$ has infinitely many atoms, list them in least-code order. Otherwise let $R_0$ be the complement of the union of its finitely many atoms. This complement is nonempty: if the atoms covered $X$, then intersecting any member of $\mathcal{B}$ with each atom would show that every member is a union of those finitely many atoms, contradicting that $\mathcal{B}$ is infinite. The set $R_0$ contains no atom. Given nonempty atomless $R_n\in \mathcal{B}$, take the least listed $B$ that splits it, put $D_n:=R_n\cap B$ and $R_{n+1}:=R_n\setminus B$, and use [L3] to continue. Both new sets are nonempty by the choice of $B$.

In the first case the listed atoms are pairwise disjoint nonempty members of $\mathcal{A}$. In the second, each $D_n\subseteq R_n$ is nonempty, $R_{n+1}$ is disjoint from $D_n$, and all later $D_m$ lie in $R_{n+1}$; hence the $D_n$ are pairwise disjoint members of $\mathcal{A}$. This constructs the required sequence without a choice principle.