Let $F = \{A\subseteq \mathbb{R}: A\subseteq [0,1] \text{ or } A^c\subseteq [0,1]\}$, is F a ring? Let $S = \mathbb{R}$ and $F = \{A\subseteq \mathbb{R}: A\subseteq [0,1] \text{ or } A^c\subseteq [0,1]\}$.
Exercise: Is $F$ a ring?
I know that $F$ is a ring if
i) $\emptyset \in F.$
ii) $A,B \in F \Rightarrow B\backslash A \in F.$
iii) $n\in \mathbb{N}, \, A_1, A_2, ..., A_n \in F \Rightarrow \, \bigcup\limits_{j = 1}^{n} A_j \in F$.
My approach: 
i) Is clearly satisfied.
ii) Pick two random subsets $A,B \in F$. We have that $A\subseteq [0,1]$ or $A^c\subseteq [0,1]$ and $B\subseteq [0,1]$ or $B^c\subset [0,1]$. Consider four different cases:
When $A,B\subseteq [0,1]$ we have that $B\backslash A = \emptyset$ or $B\backslash A = [0,a)\cup (b,1]$ with $a< b,a,b\in(0,1)$ or $B\backslash A = [a,b]$ with $a,b\in[0,1]$. In this case for every $x\in(B\backslash A)$ we have that $x\in F$
When $A \subseteq [0,1]$ and $B^c \subseteq [0,1]$ we have that $B\backslash A = B$. $B\in F \Rightarrow B\backslash A \in F$.
When $A^c \subseteq [0,1]$ and $B\subseteq [0,1]$ we have that $B\backslash A = B$. $B \in F \Rightarrow B\backslash A\in F$.
When $A^c, B^c\subseteq [0,1]$ we have that $B\backslash A = \emptyset$ or $B\backslash A \subseteq (a, b) \cup (c, d)$ with $a<b<c<d, a,b,c,d \in (-\infty, \infty)$ or $B\backslash A \subseteq [a,b]$ with $a<b, a,b \in (-\infty, \infty)$. In this case for every $x\in B\backslash A$ we have that $x\in F$.
iii) Pick any $n\in\mathbb{N}$, such that $A_1, A_2, ..., A_n \in F$. For every $1\leq i\leq n$ we have that if $x\in A_i$, then $x\in F$. Obviously for any $x\in \bigcup\limits_{i = 1}^n A_i$ we have that $x\in F$.
Question: is my solution correct/how would I solve this exercise correctly if not? Sometimes I feel my approach is a bit too intuitive (especially in part iii).
 A: Concerns ii)
$$A\subseteq[0,1]\text{ or }A^{\complement}\subseteq[0,1]\iff A^{\complement}\subseteq[0,1]\text{ or }A\subseteq[0,1]$$ or equivalently
$$A\in F\iff A^{\complement}\in F$$
So $F$ is closed under complements.
After this observation proving that $A,B\in F\implies A\cup B\in F$ is enough. 
This because:$$B\setminus A=\left(B^{\complement}\cup A\right)^{\complement}$$

Concerns iii)
we discern the following cases:
If $A,B\subseteq[0,1]$ then $A\cup B\subseteq[0,1]$
If $A^{\complement}\subseteq[0,1]$ then $\left(A\cup B\right)^{\complement}\subseteq A^{\complement}\subseteq[0,1]$
If $B^{\complement}\subseteq[0,1]$ then $\left(A\cup B\right)^{\complement}\subseteq B^{\complement}\subseteq[0,1]$ (essentially the same as the former case)

edit to clarify
Observe that: $$A=\left(A^{\complement} \right)^{\complement}\tag1$$
So in this context for $A\subseteq\mathbb R$ the following statements are evidently equivalent:


*

*$A\in F$

*$A\subseteq[0,1]\text{ or }A^{\complement}\subseteq[0,1]$ (on base of definition of $F$) 

*$A^{\complement}\subseteq[0,1]\text{ or }A\subseteq[0,1]$ (essentially the same as former statement)

*$A^{\complement}\subseteq[0,1]\text{ or }\left(A^{\complement} \right)^{\complement}\subseteq[0,1]$ (on base of $(1)$)

*$A^{\complement}\in F$ (on base of definition of $F$)

A: Your proof for iii) is not complete.
You only showed that if $A,B \subset [0,1]$, then $A\cup B \subset [0,1]$.
Similarly it is easy to show that if $A, B \subset [0,1]^C$, then $A\cup B \subset [0,1]^C$.  
Now assume $A, B \in F$ with $A \subset [0,1]$ and $B^C \subset [0,1]$.
Then $(A\cup B)^C = A^C \cap B^C \subset B^C \subset [0,1]$, so $A \cup B \in F$.

Regarding ii):
Your proof seems right, but a bit cumbersome.
I would prefer to use that $A\setminus B = A \cap B^C\subset A, B^C$.  
So if $A$ or $B^C$ are subsets of $[0,1]$, then so is $A\setminus B$.
The missing case is that $A^C$ and $B$ are subsets of $[0,1]$. Then look at the complement $(A \setminus B)^C = A^C \cup B \subset [0,1]. $
