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](https://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice/). 


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:



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> 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?
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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.)



Countably generated $\sigma$-algebras of ${\mathcal P}({\mathbb R})$ and choice
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.




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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.


