Set-theoretic equality 
Let $A⊂U^{*},B⊂U.$ Find the set $X⊂U,$ that satisfies the equation.
  $$(\overline{X \cup A}) \cup (X \cup \overline{A}) =B.$$


My thoughts:
$$\begin{align}B&=(\overline{X \cup A}) \cup (X \cup \overline{A})\\ & =  \overline{A} \cap \overline{X} \cup \overline{A} \cup X \\ &= \overline{A} \cup X.\end{align}$$
The last condition implies that  $\overline{A} \subset B$, and than $A∪B=U.$ Otherwise, the equation has no solution. If the last condition is satisfied, then suitable any set $X$, concluded between $A∩B$ и $B$. So, I came to the
necessity and sufficiency condition $$A∩B \subset X \subset B.$$
Autor's answer is $X = \overline{B}$. I think it's mistake. Can anyone explain that?

*$U$ — universal set.
 A: In your post you only proved that $B = \overline A \cup X \implies A\cap B \subseteq X \subseteq B$, although you're missing the parenthesis in the expression $(\overline A \cap \overline X)\cup \overline A \cup X$.
After parenthesizing it, your equality still holds:
\begin{align} (\overline A \cap \overline X)\cup \overline A \cup X &= [\overbrace{(\overline A\cup \overline A)}^{\overline A}\cap \overbrace{(\overline X \cup \overline A)}^{\;\;\supseteq \ \overline A}]\cup X \\ 
&= \overline A \cup X
\end{align}
Now, you know that $B = \overline A \cup X$. You can derive the inclusions from this:
$$  B = \overline A \cup X \implies X \subseteq B \text{ and } \overline A \subseteq B $$
\begin{align}
B = \overline A \cup X \implies A \cap B &= A \cap (\overline A  \cup X ) \\ 
&= \underbrace{(A \cap \overline A)}_{\varnothing} \cup (A \cap X) \\
&= A \cap X
\end{align}
$$ A \cap B = A \cap X \subseteq X \implies A \cap B \subseteq X$$
Hence, $B = X \cup \overline A \implies A \cap B \subseteq X \subseteq B$.

The converse only holds if you assume first that $\overline A \subseteq B$:
Let $U=\{1,2,3,4,5\}, A = \{1,2\}, B = \{2,3,4\}$.
Then $A \cap B = \{2\}$, so we can take $X = \{2,4\}$.
This $X$ fulfills the requirement that $A \cap B \subseteq X \subseteq B$, but $B \neq \overline A \cup X = \{3,4,5\} \cup \{2,4\}=\{2,3,4,5\}$, because $5 \notin B$.

Now, with that assumption, let's prove the converse: If $\overline A \subseteq B$ and $X$ is a set such that $A \cap B \subseteq X \subseteq B$, then $B = X \cup \overline A$
$(\subseteq)$: Let $b \in B$. 
If $b \notin A$, then $b \in \overline A$, so $b \in X \cup \overline A$.
If $b \in A$, then, since $b \in B$ too,  $b \in B \cap A \subseteq X \cup \overline A$
$(\supseteq)$: Suppose $c \in X \cup \overline A$.
If $c \in \overline A$, then $c \in B$, because $\overline A \subseteq B$ by assumption.
If $c \notin \overline A$, then $c \in X$, because $c$ is in the union, and since $X \subseteq B$ we have $c \in B$. 
