Confusion regarding the proof for "every regular lindelöf space is normal".

Question

"General Topology" by Pervin gives the following proof:

Let $A$ and $B$ be two disjoint closed sets in regular Lindelöf space $X$. Then they too are lindelöf. For every $a\in A$, there exists an open set $G(a)$ containing $a$ such that $G(a)\cap B=\emptyset$. $\bigcup_{a \in A} G(a)$ will form a cover for $A$, which will have a countable subcover as $A$ is Lindelöf. Let the countable subcover be $\bigcup_{n\in\Bbb{N}} G(a_{n})$.

Similarly, $\bigcup_{b \in B} G(b)$ will have a countable subcover $\bigcup_{n\in\Bbb{N}} G(b_{n})$ for $B$.

Then $\bigcup_{n\in\Bbb{N}}[G(a_{n})\setminus \bigcup_{i\leq n}c(G(b_{i})]$ and $\bigcup_{n\in\Bbb{N}}[G(b_{n})\setminus \bigcup_{i\leq n}c(G(a_{i})]$ will be two disjoint open sets containig $A$ and $B$ respectively.

1. I have a doubt regarding the last line. Why did we have to construct disjoint closed sets this way? Wouldn't $\bigcup_{n\in\Bbb{N}} G(a_{n})\setminus \bigcup_{n\in\Bbb{N}} c(G(b_{n}))$ and $\bigcup_{n\in\Bbb{N}} G(b_{n})\setminus \bigcup_{n\in\Bbb{N}} c(G(a_{n}))$ also be disjoint open sets containing $A$ and $B$?

2. Why do we require the regular space to be lindelöf for it to be normal? We can just take two closed sets $A$ and $B$, and determine $\bigcup_{n\in\Bbb{N}} G(a_{n})\setminus \bigcup_{n\in\Bbb{N}} c(G(b_{n}))$ and $\bigcup_{n\in\Bbb{N}} G(b_{n})\setminus \bigcup_{n\in\Bbb{N}} c(G(a_{n}))$! Is it because unless the number of open sets is countable, the notation $\bigcup G(a)$ as the cover of $A$ is invalid?

EDIT- What about the sets $\bigcup_{n\in\Bbb{N}} G(a_{n})\setminus c(\bigcup_{n\in\Bbb{N}} (G(b_{n})))$ and $\bigcup_{n\in\Bbb{N}} G(b_{n})\setminus c(\bigcup_{n\in\Bbb{N}} (G(a_{n})))$? Are they not disjoint open sets containing $A$ and $B$?

Answer 1

$\bigcup_{n \in \mathbb{N}}c(G(b_{n}))$ is an infinite union of closed sets, so it may not be closed. So when we subtract it from $G(a)$, the resulting set may not be open. However $\bigcup_{i \leq n}c(G(b_{i}))$ is closed.

Normality is in general stronger than regularity. See this question.

Answer 2

In 1. we are constructing disjoint open subsets, not disjoint closed subsets (as required for the definition of normality). An infinite union of closed sets need not be closed (e.g.: otherwise all subsets of a countable $T_1$ space like $\mathbb{Q}$ would be closed, which is false), but finite unions are. So the set is constructed as the union of (countably many) differences of open an open set with a closed set.

When $O$ is open and $F$ is closed, $O \setminus F = O \cap (X \setminus F)$, which is the intersection of two open sets, is also open. So the result of the difference of an open set and a closed set is indeed open. So the sets we construct are indeed unions of open sets, hence open. We cannot substract the union of all closed sets, as it is likely not even closed, so we do not make an open set this way.

In 2. you basically ask why the countability is so important. Well, in that case all initial segments are indeed finite: from $G(a_n)$ we substract all (finitely many!) closures of $G(b_i)$, where $i \le n$. If we would have a larger family of sets, we could try to well-order the indices as well, but at some stage we would substract unions of infinitely many closed sets again, and there goes the proof.... It's not because we can only take unions of countably many sets in general. And of course, there are spaces that are regular but not normal, so the proof must require some extra assumptions.