Inclusion and union with the complement

$\Rightarrow$. Suppose $A\subseteq B$ and let $x\in E$. If $x\in B$, then $x\in\bar A\cup B$. If $x\notin B$, then $x\notin A$ (since $A\subseteq B$), so $x\in\bar A\subseteq\bar A\cup B$. Hence every element of $E$ is in $\bar A\cup B$.

$\Leftarrow$. Suppose $\bar A\cup B=E$. If $A\not\subseteq B$, there is $x\in A$ with $x\notin B$. Then $x\notin\bar A$ and $x\notin B$, so $x\notin\bar A\cup B$, contradicting $\bar A\cup B=E$. Hence $A\subseteq B$.

Conclusion. The equivalence holds.