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.
Part of the argument is simply wrong: the base constructed is not countable, because it uses all $\epsilon>0$ instead of a countable set of $\epsilon>0$ containing arbitrarily small members, like $\{2^{-n}:n\in\Bbb N\}$ or $\left\{\frac1n:n\in\Bbb Z^+\right\}$. The intended argument could be made much more clearly. Let me simply do a more intelligible version, rather than comment specifically on this one.
For each $n\in\Bbb N$, $\{B(x,2^{-n}:x\in X\}$ is an open cover of the Lindelöf space $X$, so it has a countable subcover $\mathscr{B}_n$. Let $\mathscr{B}=\bigcup_{n\in\Bbb N}\mathscr{B}_n$; $\mathscr{B}$ is a countable family of open subsets of $X$, and I claim that it’s a base for the topology of $X$.
To see this, let $U$ be any non-empty open set in $X$, and fix $x\in U$; we must show that there is some $B\in\mathscr{B}$ such that $x\in B\subseteq U$. Since $x\in U$, and $U$ is open, we know that there is an $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$. Choose $n\in\Bbb N$ large enough so that $2^{-n}<\frac{\epsilon}2$. Now $\mathscr{B}_n$ covers $X$, so there is a $B(y,2^{-n})\in\mathscr{B}_n$ such that $x\in B(y,2^{-n})$; I claim that $B(y,2^{-n})\subseteq U$.
Suppose that $z\in B(y,2^{-n})$; then $d(z,y)<2^{-n}<\frac{\epsilon}2$. We also know that $d(x,y)<2^{-n}<\frac{\epsilon}2$, so by the triangle inequality we have $d(x,z)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$, $z\in B(x,\epsilon)\subseteq U$, and hence $B(y,2^{-n})\subseteq U$, as claimed. It follows that $\mathscr{B}$ is indeed a base for the topology on $X$.
Added: Getting back to your specific questions about the original form of the argument, I think that what you’ve not realized is that for each $\epsilon>0$ there is only one family $\{B(x_{\epsilon,n},\epsilon:n\in\Bbb N\}$ of $\epsilon$ balls being considered; there is not a separate one for each $B(y,d)$. (The double subscript on $x_{\epsilon,n}$ is necessary, because for each $\epsilon>0$ there is potentially a different $x_n$.) We use the Lindelöf property to find these countable covers of $X$ at the beginning; we don’t find new ones for each $B(y,d)$. Part of the problem, I think, is that the argument given does not adequately explain why we can be certain that there is a $B(x_{\epsilon,n},\epsilon)$ such that $d(x_n,y)+\epsilon<d$ and $\epsilon\ge d(x_{\epsilon,n},y)$.
Best Answer
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 $X$