Countably generated $\sigma$-algebras of ${\mathcal P}({\mathbb R})$ and choice It is consistent with ${\sf ZF}$ that the reals are the countable union of countable sets. Since any countable set is Borel, it follows that in any such pathological universe, let's call it $W$, every set is Borel; this has come out here before. 
Since there is a countable basis for the Borel sets, it follows that in $W$ there is a countable family of sets of reals, such that the $\sigma$-algebra they generate is all of ${\mathcal P}({\mathbb R})$. I would like to see whether we can improve this slightly:

Suppose ${\mathbb R}=\bigcup_n A_n$, where each $A_n$ is countable. Can we further assert that, in addition, the $\sigma$-algebra the $A_n$ generate is ${\mathcal P}({\mathbb R})$? More precisely, can we modify the $A_n$ to a new family $A_n'$ of countable sets with union ${\mathbb R}$ and this generating property?

Asaf's answer shows I'm being (somewhat) sloppy in this formulation. So, does it make sense to ask for some intrinsic construction? (Something other than: We start with such and such countable family that generates the Borel sets, we code it into bits of the $A_n$, and then we are done. I confess I do not see immediately how to formalize "intrinsic", perhaps it makes no sense.)
 A: We can cheat in the following way:
Suppose $\lbrace A_n\rbrace$ is a countable collection of countable sets, let $\lbrace B_n\rbrace$ be an enumeration of the countable collection of open intervals with rational end points.
Now consider the family $\lbrace A_n\cap B_m\mid n,m\in\omega\rbrace$. It is a countable family of countable sets, and every open interval $B_k$ is the countable union $\bigcup_n(A_n\cap B_k)$. Therefore the $\sigma$-algebra generated by this family is the entire power set of the real numbers.

The reason we cannot talk much about the general case is that given a countable family of countable sets $\lbrace A_n\rbrace$ we can make them disjoint via $A'_n=A_n\setminus\bigcup_{k < n} A_k$. The $A'_n$ are still countable, disjoint and have the same union - namely $\mathbb R$.
Now the $\sigma$-algebra the disjoint family generates is exactly $\lbrace\bigcup_{i\in I} A'_i\mid I\subseteq\omega\rbrace$. Since there can only be countably many singletons in this family we cannot separate all the points of $\mathbb R$ from one another and therefore we cannot generate $\mathcal P(\mathbb R)$.
So if the $A_n$ were already disjoint we could not have generated $\mathcal P(\mathbb R)$ without some sort of modification as in the first part.
