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.
For events in probability theory, Wikipedia calls events jointly or collectively exhaustive if their union is everything.
So the dual notion that their intersection is empty could be called "jointly disjoint" or "collectively disjoint".
Related question with terminology: Confusion on pairwise disjoint and disjoint
Best Answer
Maybe something along the lines of
if $A\subset B'$: $$ A\cap B' = A $$ therefore $$ A\cap B = \left(A \cap B'\right)\cap B = A \cap \left(B'\cap B\right) = A \cap \emptyset = \emptyset $$
if $B' \subset A$ then there $$ A\cap S = A = A\cap(B'\cup B) = (A \cap B') \cup (A \cap B) = B' \cup (A \cap B) $$ if $A \cap B = \emptyset$ then $A = B'$ which contradicts the first statement of $B' \subset A$
for $A = B'$ it is easy to sure that this disjoint.