Understanding the proof that regular second countable space is normal 
10.3.30 Lemma. Every regular second countable space $(X,\tau)$ is normal.
Proof. Let $A$ and $B$ be disjoint closed subsets of $(X,\tau)$ and $\mathcal{B}$ a countable basis for $\tau$. As $(X,\tau)$ is regular and $X\setminus B$ is an open set, for each $a\in A$ there exists a $V_a\in\mathcal{B}$ such that $\overline{V}_a\subseteq X\setminus B$.
As $\mathcal{B}$ is countable we can list the members $\{V_a:\ a\in A\}$ so obtained by $V_i$, $i\in\mathbb{N}$; that is, $A\subseteq\bigcup_{i=1}^\infty V_i$ and $\overline{V}_i\cap B=\varnothing$, for all $i\in\mathbb{N}$.
Similarly we can find sets $U_i$ in $\mathcal{B}$, $i\in\mathbb{N}$, such that $B\subseteq\bigcup_{i=1}^\infty U_i$ and $\overline{U}_i\cap A=\varnothing$, for all $i\in\mathbb{N}$.
Now define $U_1'=U_1\setminus\overline{V}_1$ and $V_1'=V_1\setminus\overline{U}_1$.
So $U_1'\cap V_1'=\varnothing$, $U_1'\in\tau$, $V_1'\in\tau$, $U_1'\cap B=U_1\cap B$, and $V_1'\cap A=V_1\cap A$.
Then we inductively define
$$\overline{U}_n'=U_n\setminus\bigcup_{i=1}^n\overline{V}_i,\ \text{and}\ V_n'=V_n\setminus\bigcup_{i=1}^n\overline{U}_i.$$
So that $U_n'\in\tau$, $V_n'\in\tau$, $U_n'\cap B=U_n\cap B$, and $V_n'\cap A=V_n\cap A$.
Now put $U=\bigcup_{n=1}^\infty U_n'$ and $V=\bigcup_{n=1}^\infty V_n'$.
Then $U\cap V=\varnothing$, $U\in\tau$, $V\in\tau$, $A\subseteq V$, and $B\subseteq U$.
Hence $(X,\tau)$ is a normal space. $\qquad\square$

I am trying to understand this proof. The first half was fine until we got to the part:
"Now define $U_{i}^{'}=U_i\backslash\overline{V_i}$". What is the purpose of this? How do we know each $U_{i}^{'}$ is still open after removing certain elements?
Edit: I have made some progress. Now I am stuck on the last paragraph starting with "Then we inductively define...."
Edit2 : There has got to be a typo right? There is no reason why we are defining the closure of $U'$ and only defining $V'$
 A: First of all: if $O$ is open and $C$ is closed, then $U\setminus C = U \cap (X\setminus C)$ is open as the intersection of two open sets. So that is why $U'_1$ and $V'_1$ are open, as differences of open and closed sets, and the same holds for the "inductively" defined $U'_n$ and $V'_n$: there the subtracted set is a finite union of closures of sets, so a finite  union of closed sets and hence closed. So again $$U'_n = U_n \setminus \bigcup_{i=1}^n \overline{V_i}$$ is the difference between an open set and a closed set.
The $\overline{U'_n}$ in (definition in) the text is a typo, we are just defining the open set $U'_n$, which is a "thinning down" of $U_n$. Also it's not an inductive definition at all because we define it directly from the $U_i$ and $V_i$ we already have after the first step; we don't use $U'_{n-1}$ to define $U'_n$, which would have made it a recursive definition in my preferred terminology (not inductive, which is a proof technique, in my book).
It's not too hard to reason directly why $U$ and $V$ as defined are as required:
Suppose that $x \in U \cap V$. So $x$ is in some $U'_i$ for $i \in \Bbb N$. Also $x \in V'_j$ for some $j \in \Bbb N$.
Say that $i \le j$ (the other case is similar). Then $x \in U_i$ (definition of $U'_i$), so also in $\overline{U_i}$ and $x \in V_j$ and $x \notin \overline{U_i}$ because $i \le j$ so $\overline{U_i} \subseteq \bigcup_{i=1}^j \overline{U_i}$, the set we subtract from $V_j$ to define $V'_j$. This is a contradiction and we get another contradiction if we assume $j \ge i$ (check this, it's a symmetrical argument!). So $U \cap V = \emptyset$.
To see that $B \subseteq U$ we note that the $U_i, i \in \Bbb N$ form an open cover of $B$: if $x \in B$ for some $i$ we have $x \in U_i$ (it's named as some $U_x$ first but then "renumbered"), so fix $m$ such that $x \in U_m$. Also by construction we have that all $\overline{V_i} \cap B  =\emptyset$ whatever $i$ is. Hence $x \in U_m$, and $x \notin \bigcup_{i=i}^m \overline{V_i}$ so by definition $x \in U'_m \subseteq U$.
That $A \subseteq V$ is entirely a symmetric argument again. We don't really need all the equations claimed in the proof. Presonally I would have started defining $U_i$ around $A$ and then $V_i$ around $B$, but that's a free choice. We repeatedly apply regularity for $a \in A$ and $B$ and thin out using the countable base, and then vice versa for $b \in B$ and $A$ and then refine the sets to make them disjoint in the union.
It's an instructive proof in how we can strengthen separation axioms by more global axioms like having a countable base. We cannot go fom Hausdorff to regular here, BTW. The standard example of a a Hausdorff non-regular space (that Morris also gives in an example/exercise) is second countable too. We can add paracompactness (introduced later, maybe) or compactness to be able to go from Hausdorff to normal.
