Union of sets with measure null difference 
Let $A$ be some open subset of the real line, and let $S$ be the set
  $$S = \{s \subseteq \mathbb R \mid \text{$s$ is open} \land \text{$(A\setminus s) \cup (s \setminus A)$ has null measure}\}.$$
  The problem is if the union of all elements in $S$ is an element of $S$.

I ran into this problem during a discussion about a construction involving equivalence classes based on the relation "the difference has null measure" on open subsets of the real line.
The discussion was regarding whether one can find some sort of "canonical" element in the equivalence class with the "fewest holes as possible". The suggestion was to simply take the union as shown above, but the question is if this always remains in the equivalence class, even though the union is infinite.
Of course, the requirement that $s$ is open is important here, because otherwise the union of $S$ would be $\mathbb R$, since any remaining real number $r$ would be in $A \cup \{r\}$, but luckily these sets are not open when $r \notin A$.
 A: Yes. If $S$ is a collection of open subsets of $\mathbb R$, then the union of all elements of $S$ is the union of countably many elements of $S$. Namely, let $\{I_1,I_2,\dots\}$ be the set of all rational open intervals $I$ such that $I\subseteq s$ for some $s\in S$, and for each $n$ choose some $s_n\in S$ with $I_n\subseteq s_n$. Since each element of $S$ is a union of rational open intervals, it is clear that
$$\bigcup S\subseteq\bigcup_{n=1}^\infty I_n\subseteq\bigcup_{n=1}^\infty s_n\subseteq\bigcup S.$$
(In general topological terms, the real line, being second countable, is hereditarily Lindelöf.)
Now, using the countable additivity of Lebesgue measure, it should be easy to prove that $\bigcup_{n=1}^\infty s_n\in S$.
A: As an example: taking $A  =(0,1) \cup (1,2)$ the union of your $S$ will be $(0,2)$ I think. 
Intuitively:
$A$ can be written uniquely as a disjoint union of open intervals / segments (the segments can be of the form (-\infty, a)$ or $(a,+\infty)$ or $(-\infty,+\infty); of course only one of them can occur or one of each of the first two types). The union of the $S$ for this $A$ will be the same $A$ unless we have a situation that e.g. we have an interval $(a,b)$ and $(c,d)$ in the unique decomposition with $b=c$, in which case we can add this point and get $(a,d)$ as part of the union of $S$. 
So consider the isolated points of $\mathbb{R}\setminus A$, call the set of these $I$ and then show that $\bigcup S = A \cup I$. I think this holds.
added Due to an example by the OP in the comments below, I think $I$ should be adapted to be not just the isolated points but the scattered (non-perfect) part of $\mathbb{R}\setminus A$.
