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.
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.
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.