$$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B)\tag{1}$$
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 $(x \in \overline{A} \cap B)$ ....
Best Answer
It’s a little easier to go the other way: $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$, and you’d like to show that this equals $(A\cap B)\cup C$ if and only if $C\subseteq A$.
Your Venn diagrams show what happens when $C\subseteq A$, so they’re useful in proving one direction of the desired result: if $C\subseteq A$, then the two sets are equal. To see how you might prove the other direction, i.e., that if $C\nsubseteq A$, then the two sets are not equal, you’d do better to look at a Venn diagram showing $C\nsubseteq A$.