Prove that $A=B$ Let A and B be two subsets of some universal set.
Prove that if $(A ∪ B)^c$ = $A^c ∪ B^c$, then $A=B$
ATTEMPT: Let y $\in (A ∪ B)^c$ this means that $y\notin A$ or $y\notin B$ which is equal to $A^c ∪ B^c$ or also  $y\notin A$ or $y\notin B$. How do I get $A=B$ though? As of now, I have two equal sides:
$y\notin A$ or $y\notin B =  y\notin A$ or $y\notin B$
 A: If you’re going to try to prove directly that $A=B$, you don’t want to start with some $x\in(A\cup B)^c$: you want to start with some $x\in A$ and show that it’s in $B$, and vice versa. (Incidentally, $x\in(A\cup B)^c$ means that $x\notin A\cup B$, which means that $x\notin A$ and $x\notin B$, not that $x\notin A$ or $x\notin B$.)
Suppose that $x\in A$. Then $x\in A\cup B$, so $x\notin(A\cup B)^c$. By hypothesis $(A\cup B)^c=A^c\cup B^c$, so $x\notin A^c\cup B^c$. And $A^c\cup B^c=(A\cap B)^c$ by one of the de Morgan laws, so $x\notin(A\cap B)^c$, and therefore $x\in A\cap B\subseteq B$. Thus, $A\subseteq B$. The reverse inclusion can be proved in exactly the same fashion.
You could also start by applying the de Morgan law: the hypothesis is equivalent to $(A\cup B)^c=(A\cap B)^c$ and hence to $A\cup B=A\cap B$. From this it’s very easy to show that $A\subseteq B$ and $B\subseteq A$, especially if you know about the symmetric difference, which I’ll write $\triangle$:
$$(A\setminus B)\cup(B\setminus A)=A\mathbin{\triangle}B=(A\cup B)\setminus(A\cap B)=\varnothing\;.$$
A: Suppose that $A\neq B$.  Then, without loss, we may choose $x\in A\cap B^c$.  Since $x\in B^c$, we have $x\in A^c\cup B^c$.  However, also $x\in A$ so $x\in A\cup B$ and therefore $x\notin (A\cup B)^c$.  Hence $(A\cup B)^c\neq A^c\cup B^c$, and we've proved the contrapositive.
On the other hand, if $A=B$, then $A\cup B=A$ so the LHS is $A^c$, while the RHS is $A^c\cup A^c=A^c$.  Thus the statement is actually true in both directions.
A: You don't want to be setting statements equal to each other here. Start with something you know, or assume, and show what you want.
Let $a \in A$. Since $a \in A \cup B$, we know $a \notin (A \cup B)^c$, and also $a \notin A^c \cup B^c$. This tells us that $a$ can't be in $B^c$. And therefore $a \in B$.
What does this tell us? Can you do the other half?
A: Let $x\in A$, then clearly 
$$x\not\in (A\cup B)^{c} = A^{c} \cup B^{c}$$
It follows that
$$x\notin A^{c} \text{ and } x\not\in B^{c}$$
if $x\not\in B^{c}$, then $x\in (B^{c})^{c} = B$. So $A\subseteq B$. Repeat for $x\in B$.
