Proving with subsets Confused at how to prove this.

An Attempt:
x ∈ (A∪B)c
i.e x ∉ A or x ∉ B
x ∈ Ac or x ∈ Bc
x ∈ Ac ∪ Bc
I don't know how to go as far as to prove that if statement leads to A = B.
Thought of using proof by contradiction but not sure where A ≠ B will lead me. Pretty confused right now and would appreciate assistance.
 A: Let $a \in A$, we want to show that $a\in B$ as well. Suppose not to get a contradiction, i.e. $b\in B^c$. 
First note that  since $a\in A$ we must have  $a \in A \cup B$, therefore $a \notin (A \cup B)^c $. However since $a \in B^c$ we have  $a \in A^c \cup B^c$.  But this constitutes a contradiction to the fact $(A \cup B)^c =  A^c \cup B^c $.
A: (1)...$x\in (A\cup B)^c\iff$ $  x\not \in A\cup B\iff$ $ (x\not \in A\land x\not \in B)\iff$ $ (x\in A^c\land x\in B^c)\iff$ $\iff x\in A^c\cap B^c.$ $$\text {So for all }A,B \text { we have } (A\cup B)^c=A^c\cap B^c.$$ $$\text  {Hence, if } (A\cup B)^c=A^c\cup B^c \text  { then } A^c\cap B^c=A^c\cup B^c.$$
(2)... For any sets $X, Y$ we have : $[X\cap Y]=[X\cup Y]\implies$ $$ \implies X\subset [X\cup Y]=[X\cap Y]\subset Y\subset [Y\cup X]=[Y\cap X]\subset X\implies$$ $$\implies   X\subset Y\subset X \implies X=Y.$$ 
(3)...Therefore $(A\cup B)^c=A^c\cup B^c\implies A^c\cap B^c=A^c\cup B^c\implies A^c=B^c\implies$ $\implies A=(A^c)^c=(B^c)^c=B.$
A: Using $(A\cup B)^c=A^c\cup B^c \iff A^c\cap B^c=A^c\cup B^c$, we have
\begin{align}
A&=(A^c)^c \\
&=(A^c\cup(A^c\cap B^c))^c \\
&=(A^c\cup(A^c\cup B^c))^c \\
&=(A^c\cup B^c)^c \\
&=((A^c\cup B^c)\cup B^c)^c \\
&=((A^c\cap B^c)\cup B^c)^c \\
&=(B^c)^c \\
&= B\ .
\end{align}
