Proving that $A\cap B \subseteq C \iff A \subseteq \overline{B} \cup C$

I'm trying to prove the following statement:

$$A\cap B \subseteq C \iff A \subseteq \overline{B} \cup C$$

I need to do it using a formal proof.. I've tried to do it for some time now and couldn't find anything close..

Thanks!

$\Rightarrow$:

$$A = \underbrace{(A\cap B)}_{\subseteq C} \cup \underbrace{(A\cap\overline{B})}_{\subseteq \overline{B}} \subseteq C\cup \overline{B}$$

$\Leftarrow$:

$$A\cap B \subseteq (\overline{B}\cup C)\cap B = (\overline{B}\cap B)\cup (C\cap B) = C\cap B\subseteq C$$

I suppose that $\bar B$ is the complementary of $B$.

Suppose $A\cap B\subset C$ and let $x\in A$. If $x\in A\cap B$, by hypothesis, $x\in C$ and thus $x\in \bar B\cup C$. If $x\notin A\cap B$, then $x\in A\cap \bar B$ and thus $x\in \bar B$. Therefore $x\in \bar B\cup C$.

Reciprocally, suppose that $A\subset \bar B\cup C$ and let $x\in A\cap B$. In particular, $x\in A$ and $x\in B$. By hypothesis, $x\in \bar B\cup C$. If $x\in \bar B$, we have a contradiction with $x\in B$, therefore $x\in C$, what prove the claim.