Prove that: For all sets $A$ and $B$, $A\cap B = A \cup B\Leftrightarrow A = B$. I wasn't able to prove this statement. I'd much appreciate if you could lend a helping hand.
 A: Prove the contrapositive.  Assume $A\ne B$.
Then without loss of generality there is $x\in A$ such that $x\not\in B$.
(If not, switch $A$ and $B$.)
Then $x\in A\cup B$ but $x\not\in A\cap B$.

Addendum in response to comment where OP asked for direct proof:
If $A=B$, then $A\cup B=A\cup A=A=A\cap A=A\cap B$.
Conversely, if $A\cup B=A\cap B$, then $x\in A\cup B\iff x\in A\cap B$,
so $x\in A $ or $x\in B\iff x\in A $ and $x\in B$,
so $x\in A\implies x\in A $ or $x\in B \implies x\in A $ and $ x\in B\implies x\in B$,
and likewise $x\in B\implies x\in A$, so $A=B$.
A: We know that
$$A\cap B\subset A \subset A\cup B$$
and
$$A\cap B\subset B\subset A\cup B$$
So,
$$A\cup B =A\cap B \implies $$
$$A\cup B=A\cap B = A =B$$
Now,
$$A=B \implies A\cup B=A=A\cap B$$
A: We have $A\cup B=(A\cap B)\cup(A\setminus B)\cup(B\setminus A)$, with all the components on the RHS disjoint.
So if $A\cap B=A\cup B$, then $A\setminus B=\emptyset$ and $B\setminus A=\emptyset$.
But $A\setminus B=\emptyset\iff A\subseteq B$.
Therefore we have  $A\subseteq B, B\subseteq A$, therefore $A=B$.
