Simple set proof (is it right?) Statement: $A\cup B = A\cap B \iff A=B$
Proof: For the converse assume that $A=B$. We need to show that $A\cup B \subseteq A\cap B$ and $A\cap B \subseteq A\cup B$. First we show that $A\cup B \subseteq A\cap B$. Let $x \in A \cup B$, then $x \in A$ or $x \in B$. Since $A=B$ it follows that $x \in A$ and $x \in B$. So $x \in A\cap B$ and thus $A\cup B \subseteq A\cap B$. Now for the other way around. Let $x \in A\cap B$, then $x \in A$ and $x\in B$. So $x \in A\cup B$ (because it still holds when $x \in A$ and $x \in B$). Thus $A\cap B \subseteq A\cup B$ and now we have proved that $A\cup B = A\cap B$.
Assume that $A\cup B = A\cap B$. We need to show that $A\subseteq B$ and $B\subseteq A$.
First we show that $A\subseteq B$. Let $x \in A$ , then $x \in A \cup B$. Since $A\cup B=A\cap B$ it follows that $x \in A\cap B$. So $x \in A$ and $x \in B$. Thus $A\subseteq B$. The other way is similar. End of proof.
It feels like the converse proof is a bit.. not perfect/right. Can someone verify? Thanks!
EDIT: I edited the proof. Is it (more) correct now?
 A: Assume A = B.  Then
A $\cup$ B = A $\cup$ A = A = A $\cap$ A = A $\cap$ B.  
Conversely:
A subset A $\cup$ B = A $\cap$ B subset B;
In a like fashion B subset A.  
A: When you assume $A = B$, you want to show $A\cup B = A\cap B$. That means you start with an $x\in A\cup B$ and show that $x\in A\cap B$, and vice versa. What you have done is to assume that $A = B$ and then start with $x\in A$. That will not lead where you want it to lead.
Assume $A= B$, and let $x\in A\cup B$. By definition of $\cup$ we have $x\in A$ or $x\in B$. However, we can't have one without the other, as $A = B$. Thus we must have $x\in A$ and $x\in B$, which gives us $x\in A\cap B$. This shows $A\cup B\subseteq A\cap B$. For $A\cap B\subseteq A\cup B$, we don't need the assumption at all; this is always true: Take $x\in A\cap B$. By definition of $\cup$ we have $x\in A$ and $x\in B$. By basic logic, it must then be true that $x\in A$ or $x\in B$, which gives us $x\in A\cup B$.
This shows that $A = B\implies A\cup B = A\cap B$. The other implication is somewhat similar. Note that the crucial part of assuming $A\cup B = A\cap B$ is the inclusion $A\cup B\subseteq A\cap B$, as the other inclusion is always true, and thus not something special.
