A confusion regarding proof of "Every regular second countable space is normal", given in Munkres Let $X$ a topological space that is second countable and regular. Let $C,D$ be closed disjoint subspaces of $X$. By regularity, we can find for each $c \in C$ , disjoint open sets $U_c$ and $V_c$ such that
$c \in U_c$ &   $ D \subset V_c $,
Therefore $\{U_c : c \in C\}$ is open cover of $C$ , and so (since $X$ is second countable) it has countable subcovers, say $\{U_n : n \in \mathbb{N}\}$ .
Now Let $U_n' = U_n \setminus \bigcup_{i=1}^n \overline{V_i}$, and $V_n' = V_n \setminus \bigcup_{i=1}^n \overline{U_i}$.

Define $U = \bigcup_{n=1}^{\infty} U_n'$ and $V =  \bigcup_{n=1}^{\infty} V_n'$.
Then $U$ and $V$ are open, since they are the union of open sets.

This is where I am having trouble. How can we claim that $lim_{n\to \infty} U_n' = lim_{n\to \infty} \left(U_n \setminus \bigcup_{i=1}^n \overline{V_n}\right) $ is open.
 A: Just remember that, if $X$ is a topological space, and $A, B\subset X$, then
$$\text{Cl}_X(A\cup B)=\text{Cl}_X(A)\cup \text{Cl}_X(B)$$
Then you may write $U'_n=U_n\backslash \bigcup_{i=1}^n \overline V_i = U_n \backslash \overline{\bigcup_{i=1}^nV_i} \ , \ V'_n=V_n\backslash \bigcup_{i=1}^n \overline U_i = V_n \backslash \overline{\bigcup_{i=1}^nU_i}$
And then
$$U = \bigcup_{n=1}^{\infty}U'_n=\bigcup_{n=1}^{\infty} \big(U_n \backslash \scriptsize \overline{\bigcup_{i=1}^{n}V_i} \big)$$
and similarly
$$V = \bigcup_{n=1}^{\infty}V'_n=\bigcup_{n=1}^{\infty} \big(V_n \backslash \scriptsize \overline{\bigcup_{i=1}^{n}U_i} \big)$$
Where clearly each of the sets from the infinite union is open: in the first equation we substract from each $U_n$ a closed subset, namely $\overline{\bigcup_{i=1}^{n}V_i}$, and in the second, we perform an analogous proceeding substracting from each $V_n$ the closed set $\overline{\bigcup_{i=1}^{n}U_i}$. Therefore, both $U$ and $V$ are open in $X$
A: Your problem was solved, but your construction doesn't work since you can't prove that $C$ is contained in $U$. However, you can repeat the construction of the $U_n$ in $D$ and define $V_n$ in the same way. That construction will work.
