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 $\mathcal P(\mathbb R)$.

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.

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.

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 $\mathcal P(\mathbb R)$.