Proving Regularly Closed If $U$ is open and $A=\overline{U}$, then $A$ is regularly closed.
Note that: $A=\overline{U}=U \cup U' \Rightarrow U \subset A \Rightarrow U \subset Int(A)$, since $U$ is open.
A set $A$ is regularly closed iff $A=\overline{Int(A)}.$
$\Rightarrow$
Let $x\in A=\overline{U}$. Let $V$ be an open set containing $x$. Then $V \cap U \neq \emptyset$ and since $U \subset Int(A)$, we have $V \cap Int(A) \neq \emptyset$. Then since $Int(A) \subset \overline{Int(A)}$, $x \in \overline{Int(A)}$.
Thus, $\overline{U}=A \subset \overline{Int(A)}$. 
$\Leftarrow$
Let $x \in \overline{Int(A)}$. Let $W$ be an open set containing $x$. Then $W \cap Int(A) \neq \emptyset$...
Am I incorrectly assuming in the first part of the proof that if $x \in V$, then $x \in Int(A)$?
Could I have some help on how to correct and finish this proof?

For all subsets $A$ of a space $X$ we have: (denoting interior of $A$ by $A^\circ$ and closure of $A$ by $A^-$):
$$A^{\circ-\circ-} = A^{\circ-}\tag 0$$
which can be seen as follows:
$$A^\circ \subseteq A^{\circ-}$$ because any set is a subset of its closure.
Taking the interior on both sides ($C \subseteq D$ implies $C^\circ \subseteq D^\circ$) we then get (also using $C^{\circ\circ}=C^\circ$ for all $C$):
$$A^\circ \subseteq A^{\circ-\circ}$$
and taking closures on both sides we get:
$$ A^{\circ-} \subseteq A^{\circ-\circ-}\tag 1$$
On the other hand, as any interior of a set is a subset of that set:
$$A^{\circ-\circ} \subseteq A^{\circ-}$$
and taking the closure on both sides ($C \subseteq D$ implies $C^- \subseteq D^-$) we get (also using $C^{--} = C^-$ for all $C$):
$$A^{\circ-\circ-} \subseteq A^{\circ-}\tag 2$$
Now (1) and (2) together give the equality (0).
Another dual equality is given by
$$A^{-\circ-\circ} = A^{-\circ}\tag 3$$
which has an analogous proof:
First clearly ($C \subseteq C^-$ for all $C$):
$$A^{-\circ} \subseteq A^{-\circ-}$$
taking interiors on both sides:
$$A^{-\circ} \subseteq A^{-\circ-\circ}\tag 4$$
while also clearly ($C^\circ \subseteq C$ for all $C$)
$$A^{-\circ} \subseteq A^{-}$$
taking the closure on both sides:
$$A^{-\circ-} \subseteq A^{-}$$
and taking the interior on both sides:
$$A^{-\circ-\circ} \subseteq A^{-\circ}\tag 5$$
and (5) and (4) give (3) again.
Then $A = U^-$ is regular closed as, using $U^\circ= U$ as $U$ is open:
$$A = U^{-} = U^{\circ-} = U^{\circ-\circ-}= A^{\circ-}$$
and so $A$ is regular closed.
