Prove $B=(A \cap B) \cup (B-A)$ To show this proof I understand that you have to show that B $\subset$ (A $\cap$ B)$\cup$(B-A) and also (A $\cap$ B) $\cup$ (B-A) $\subset$ B.
I start off the proof by saying: Suppose that x $\in$ B, then x  $\in$ (A $\cap$ B) and x $\in$ (B-A). 
I also know that (B-A) denotes the set of elements in B that are not in A.
 A: $(A\cap B)\cup (B\setminus A)=(A\cap B)\cup (B\cap \overline{A})$.
By the Distributive Property $=B\cap(A\cup \overline{A})$. 
By the Complement Law $=B\cap U$. By the Identity Law $=B$.
A: To start, $x \in (A \cap B) \cup (B-A)$. Think of $\cup$ like logical or; to be in the union of two sets means to be in one set or the other. First, assume $x \in B$ and $x \notin A \cap B$. This means that $x$ is either not in $A$ or not in $B$ (can you see why?). Then prove $x \in B-A$.
For the other side, you in fact need to prove $A\cap B \subset B$ and $B-A \subset B$. This is because you need to show that their union is a subset of $B$, which means that each of those two sets is subset of $B$. Don't think too hard on this one; the proof should flow naturally from what you have and intuition.
A: It sounds like you already did the first part. You showed that
$$B \subset  (A\cap B) \cup (B -A)$$
by proving the statement
$$x \in B \implies x \in (A\cap B) \cup (B - A).$$
Now to show that 
$$(A\cap B) \cup (B -A) \subset B$$
you could try to prove the statement
$$x \in (A\cap B) \cup (B -A) \implies x \in B$$ 
but I feel it is slightly easier to prove the contrapositive:
$$x \notin B \implies x \notin (A\cap B) \cup (B -A).$$
A: If $x\in B$ then you know that either $x\in A$ or $x\in A^{c}$, where $A^{c}$ denotes the complement to $A\ .$ Thus, $x \in A\cap B$ or $x \in B\cap A^{c}\ .$ But $B \cap A^{c} = B-A\ ,$ so you have shown that $$B \subset (A\cap B) \cup (B-A)\ .$$
For the converse, you can use the distributive law to manipulate the expression $(A\cap B) \cup (B\cap A^{c})\ .$ $$(A\cap B) \cup (B\cap A^{c}) = B \cap(A \cup A^{c}) = B \cap \Omega = B\ ,$$ where $\Omega$ denotes the universal set. Thus, if $x \in (A\cap B) \cup (B-A)$ then $x \in B\ .$
As you see, the discussion on the converse shows that $B =(A\cap B) \cup (B-A)$ is a consequence of a distributive law of the set-operations $\cap$ and $\cup\ ,$ so it is not neccessary to split the statement $B =(A\cap B) \cup (B-A)$ into two parts $B \subset (A\cap B) \cup (B-A)$ and $(A\cap B) \cup (B-A) \subset B\ .$
