I am trying to understand the last line of this proof. Why is the union of all $c(B_k)$ equal to $G$? I dont understand the difference between $B_{n(x),\lambda(x)}$ and $B_{k,\lambda(x)}$.
I know that there must be a member of the base containing $x$ and is contained in $G$ but why is the union of all these members is also contained in $G$.
The proof somewhat more spelled out but essentially the same:
For all $x \in G$ we have some $n(x) \in \Bbb N$ and some $\lambda(x) \in \Lambda_{n(x)}$ so that $$x \in B_{n(x), \lambda(x)} \subseteq \overline{B_{n(x), \lambda(x)}} \subseteq G\tag{1}$$ by openness of $G$, regularity of $X$ and the fact that $\{B_{n,\lambda}\mid n \in \Bbb N, \lambda \in \Lambda_n\}$ is a base for $X$. (I'll denote the closure by the overline, not by $c$, as your text does, in the following).
Now define, for each $k \in \Bbb N$ temporarily fixed $G_k = \{x \in G\mid n(x)=k\}$. It's clear by definition that $\bigcup_n G_n = G$ (each point of $x \in G$ has such an $n(x)$ and it must be one of the natural numbers).
Now with $k$ still fixed, any $x \in G_k$ (so that $n(x)=k$), the collection $\mathcal{B}_k:=\{B_{k,\lambda(x)}\mid x \in G_k\}$ is a subfamily of the locally finite family $\{B_{k,\lambda}: \lambda \in \Lambda_k\}$ and so also locally finite and "closure preserving" ( I think that fact is called 10.2.1 in your text) which means that $$\overline{\bigcup \mathcal{B}_k}=\bigcup \{\overline{B_{k,\lambda(x)}}\mid x \in G_k\}\subseteq G$$ (the last inclusion follows from $(1)$. Now we define $B_k = \bigcup \mathcal{B}_k$ as the text does, and have $\overline{B_k} \subseteq G$. But as every $x \in G$ is in some unique $G_k$ and so in $B_k$ (i.e. in its "own" $B_{k,\lambda(x)}$) and in $\overline{B_k}$ too, we get that $G=\bigcup_k \overline{B_k}$ (all unionants are subset of $G$ too, as we saw, so we have a double inclusion). Hence $G$ is an $F_\sigma$.
I hope this version helps somewhat.