Halmos Measure Theory section 5 Theorem D Theorem D: If $E$ is any class of sets, and $\mathrm{E}$ is any set is $S = S(E)$ , then there exist a countable subclass $D$ of $E$ such that $\mathrm{E}$ is in $S(D)$ . 
Proof: The union of all the $\sigma$-subrings which are generated by countable sublasses of $E$ is a $\sigma$-ring containing $E$ and contained in $S$; it is therefore identical with $S$
My confusions: (i) Why the union of all the $\sigma$-subrings is a $\sigma$-ring? The set difference isn't considered here (ii) Why this union is containing $E$?
Any help is greatly appreciated.  
 A: Following Edgar's hint, I will attempt a comprehensive proof.
Let us denote the union of all sigma sub-rings of $S$ generated by some countable subclass of $E$ by $U$ (it is a class of sets). Suppose sets $A,B\in U$. Hence, we know that $A,B$ are each in the sigma rings generated by some countable subclass of $E$, let us denote the countable subclasses each by $P,Q$. Now, as GEdgar pointed out, $S(P\cup Q)$ is also in $U$ since $P\cup Q$ is countable. Therefore, we have that $A,B\in S(P\cup Q)$ and hence $A\cup B\in U$. Similarly, we also have $U$ is closed under differences. To see $U$ is closed under countably infinite unions, we note that countable unions of countable subclasses is countable, and hence any countable unions of countable sets in $U$ is also in the sigma ring generated by a countable subclass of $E$. Therefore, $U$ is closed under countable union.  
To see why $E$ is a subclass of $U$, let us suppose the set $K\in E$. Note that the subclass $\{K\}$ is countable, so the sigma ring generated by it is also in $U$ as desired. The generated sigma ring includes $K$ by definition. Hence, $K\in U$ as desired.
Moreover, since each sigma ring generated by a subclass of $E$ is a subclass of $S(E)$, $U\subset S(E)$. Hence $U=S(E)$ since $S(E)\subset U$ by the definition that $S(E)$ is the smallest sigma ring containing $E$.
Hope this helps.
