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?

Since U is open U = int U.
So A = cl U = cl int U.  Thus
cl int A = cl int cl int U = cl int U = A.
Exercise.  Show by set inclusions cl int cl int U = cl int U. 
