$$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B)\tag{1}$$
Step (1) $ (A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap \overline{A} \cup \overline{B}$
Yes, correct, but you should keep parentheses: $ (A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap (\overline{A} \cup \overline{B})$
Step (2) Not correct: $ (A \cup B) \cap \overline{(A \cap B)}$ = $A \cup (B \cap \overline{A}) \cup \overline{B} $, by associative property.
No: the associative property (like the commutative property) applies to a chain of unions, or a chain of intersections, but not a mixed chains of unions and intersections. This is why parentheses in the first step are needed. You need to use distribution:
$ (A \cup B) \cap \overline{(A \cap B)}$ = $(A \cup B) \cap (\overline{A} \cup \overline{B}) = [(A\cup B) \cap \overline{A}] \cup [(A \cup B) \cap \overline{B}]$
Can you take it from here? You can use distributivity again...then use that fact that $A \cap \overline{A} = \varnothing$. Likewise for $B\cap \overline{B} = \varnothing$. Simplify.
Another approach is to unpack what proving your equality $(1)$ requires:
In general, we prove that for two sets $P, Q$, $$P = Q \iff P \subseteq Q \text{ AND}\;\;Q\subseteq P$$
For your equality $(1)$, that means you can prove the equality by proving:
$$[(A \cup B) \cap \overline{(A \cap B)}] \subseteq [(A \cap \overline{B}) \cup (\overline{A} \cap B)]\tag{2}$$
and by proving $$[(A \cap \overline{B}) \cup (\overline{A} \cap B)] \subseteq [(A \cup B) \cap \overline{(A \cap B)}]\tag{3}$$
Unpack what this means in terms of "chasing elements", for which I'll give you a start:
$(2)$ If $x \in [(A \cup B)\cap \overline{(A \cap B)}]$ then $x \in (A\cup B)$ AND $x \notin (A \cap B)$, which means $(x \in A$ OR $x \in B)$ AND $(x \notin A$ OR $x \notin B)$...
$(3)$ If $x \in [(A\cap \overline{B})\cup (\overline{A} \cap B)]$, then ...$(x \in A \cap \overline{B})$ OR $(x \in \overline{A} \cap B)$ ....
Like I said in my comment, I'm pretty sure that $\overline A$ is referring to the complement of $A$ in $S$. The way to prove this problem is to just blindly "chase elements":
Let $x\in\overline{A\cap B}$. Then $x\in S$ but $x\notin A\cap B$. Therefore $x\notin A$ or $x\notin B$. This precisely means $x\in\overline A\cup\overline B$, so $\overline{A\cap B}\subseteq\overline A\cup\overline B$.
I would encourage you do the other direction on your own. Just follow the same procedure I did above, and follow the definitions to show $\overline A\cup\overline B\subseteq\overline{A\cap B}$.
Best Answer
This is false if your space is $\{0\}$ and $A=B=\emptyset$. Because $\overline{A}=\overline{B}=\{0\}$, the left hand side becomes $\overline{\{0\}}\cup\overline{\emptyset}=\emptyset\cup\{0\}=\{0\}$ which is not $A$.
What is true, is that $\overline{\overline{A}\cup\overline{B}}\cup\overline{\overline{A}\cup B}=A$.
Proof in words:
If we have $\overline{V\cup W}$, this is everything not in $V\cup W$, so everything neither in $V$ nor $W$. Hence $\overline{V\cup W}=\overline{V}\cap\overline{W}$. So we get $\overline{\overline{A}\cup\overline{B}}\cup\overline{\overline{A}\cup B}=(A\cap B)\cup(A\cap\overline{B})=A\cap(B\cup\overline{B})=A$.