Proof on disjoint union of sets $A$ and $B$ Here is my attempt at a solution for a proof about disjoint unions of sets $A$ and $B$. Can you please point out the mistakes? Thank you all.
Let $A$ and $B$ be any set. Prove: 


*

*$A$ is the disjoint union of $A\setminus B$ and $A\cap B$

*$A\cup B$ is the disjoint union of $A\setminus B$, $A\cap B$, and $B\setminus A$.


Part 1
Suppose $A$ and $B$ are not disjoint. Then,
$$
\{x | x\in A\setminus B\text{ and }x\in A\cap B\}
$$
Since $A\subseteq A\setminus B$ and $A\subseteq A\cap B$, $A\setminus B = A\cap B$. So, $A\setminus B\nsubseteq B$, or $A\setminus B\subseteq B^{C}$, and $A\cap B\subseteq B$.
Then, $B\cup B^{C} = \emptyset$, and $A$ and $B$ are disjoint.
Part 2
Suppose $A\cup B$ is not disjoint. Then,
$$
\{x | x\in A\setminus B\text{ and }x\in A\cap B\text{ and }x\in B\setminus A\}
$$
$A\setminus B\subseteq B^{C}$, $A\cap B\subseteq A$, $A\cap B\subseteq B$, and $B\setminus A\subseteq A^{C}$. But since $B\cup B^{C} = \emptyset$ and $A\cap A^{C} = \emptyset$, $A\cup B$ is disjoint.
 A: In the first one, $A$ is not a subset of $A\setminus B$, but rather the other way around, that is $A\setminus B\subseteq A$ (Consider $A=\{0,1,2\}$ and $B=\{1\}$ as a counterexample to your statement).
Also $B\cap B^c=\varnothing$, and rather $B\cup B^c$ is everything.
You need to argue, however, $x\in A\setminus B$ then $x\notin B$, therefore $x\notin A\cap B$; and vice versa (that is $x\in A\cap B$ then $x\notin A\setminus B$). Then you need to show that $x\in A$ then either $x\in A\cap B$ or $x\in A\setminus B$ (which really boils down to the fact that either $x\in B$ or $x\notin B$).
In the second one the argument is completely unclear to me. Using the first part you can write $A=(A\setminus B)\cup (A\cap B)$ as a disjoint union, as well to apply the same argument on $B = (B\setminus A)\cup (B\cap A)$.
Now use the fact that $A\setminus B$ and $B\setminus A$ are disjoint to prove that the decomposition of $A\cup B=(A\setminus B)\cup(B\setminus A)\cup (A\cap B)$ is a disjoint union.
Lastly (after the $\LaTeX$ was fixed by cardinal) note that:
$$A\cup B=\{x\mid x\in A\ \mathbf{or}\  x\in B\}$$
While you wrote that this is "$x\in A\setminus B$ and $x\in A\cap B$ and $x\in B\setminus A$" which would be the intersection, which you can prove is empty.
A: You are using proof by contradiction unnecessarily. The disjointness arguments are much clearer when done directly. For example, $A \cap B \subseteq B$, and so $A \cap B$ must be disjoint from $A \setminus B$.
For the union, pick any element from $x \in A$. We want to show that it is in either $A \cap B$ or $A \setminus B$. This is a good place to use contradiction. Suppose $x$ is not in one of the sets (say $A \cap B$), since if it is, we are already happy. Since $x$ is an element of $A$ but not an element of $A \cap B$, it follows that $x$ is not an element of $B$. So, $x \in A$ and $x \notin B$, which means precisely that $x \in A \setminus B$.
A: Your statements 1. and 2. are so obvious that any "formal proof" actually weakens their credibility.
Ad 1. Any element $x\in A$ either does not belong to $B$ or is  an element of $B$ as well.
Ad 2. For any element of $x\in A\cup B$ exactly one of the following is true: $x$ belongs solely either to $A$ or to  $B$, or to both of them.
