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.

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 $X$

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.