You’re in trouble already in the first line of your argument:
Let $a\in A$ be an arbitrary element, we well show that $a\in A\cap\overline B\cap\overline C$.
You can’t show this, because it’s not necessarily true that an arbitrary element of $A$ belongs to $\overline B\cap\overline C$. It also isn’t what you want to show. At this point you’re trying to show that $$(A\setminus C)\setminus(B\setminus C)\subseteq(A\setminus B)\setminus C\;,\tag{1}$$ so you should be starting with an arbitrary $a\in(A\setminus C)\setminus(B\setminus C)$, like this:
Let $a\in(A\setminus C)\setminus(B\setminus C)$ be arbitrary. Then $a\in A\setminus C$, and $a\notin B\setminus C$. Since $a\in A\setminus C$, $a\in A$ and $a\notin C$. Since $a\notin B\setminus C$, either $a\notin B$, or $a\in C$. But we know that $a\notin C$, so it must be the case that $a\notin B$. Putting the pieces together, we see that $a\in A$ and $a\notin B$, so $a\in A\setminus B$, and moreover $a\notin C$, so $a\in(A\setminus B)\setminus C$. This proves $(1)$.
To complete the proof you must show that
$$(A\setminus B)\setminus C\subseteq(A\setminus C)\setminus(B\setminus C)\tag{2}\;,$$
so this time you should start with an arbitrary element of $(A\setminus B)\setminus C$:
Let $a\in(A\setminus B)\setminus C$ be arbitrary. Then $a\in A\setminus B$, and $a\notin C$. Since $a\in A\setminus B$, $a\in A$, and $a\notin B$. We now know that $a\in A$ and $a\notin C$, so $a\in A\setminus C$. We also know that $a\notin B$, so $a\notin B\setminus C$, and therefore $a\in(A\setminus C)\setminus(B\setminus C)$. This proves $(2)$, and $(1)$ and $(2)$ together yield the desired result that $(A\setminus C)\setminus(B\setminus C)=(A\setminus B)\setminus C$.
There’s nothing tricky about any of this: it’s all just using the definition of set difference. It’s an example of what I call a follow-your-nose proof: you do the most straightforward, natural thing at each step, and it works.
