Your claim that $\mathcal{G}$ is stable under countable intersection is not true, in general, and so the procedure you describe doesn't work. Here is a counterexample. It may be more sophisticated than necessary, but I don't know anything simpler. (However, you may also find it instructive to examine why you thought it was true, and find the error in your reasoning.)
Let $X = \{0,1\}^{\mathbb{N}}$ be Cantor space, viewed as the set of all functions $f : \mathbb{N} \to \{0,1\}$, equipped with the product topology. Recall that $X$ is a compact Hausdorff space (indeed metrizable), so it satisfies the Baire category theorem; we shall use this later.
Let $\pi_n : X \to \{0,1\}^n$ be the projection map onto the first $n$ coordinates (i.e. $\pi_n(f) = f|_{\{1,\dots,n\}}$). Let $$\mathcal{B} = \{ \pi_n^{-1}(A) : A \subseteq \{0,1\}^n, n = 1,2,\dots\}$$ be the collection of all "cylinder sets". Note that $\mathcal{B}$ is a Boolean algebra, and that every set in $\mathcal{B}$ is closed (and also open). In fact, $\sigma(\mathcal{B})$ is the Borel $\sigma$-algebra of $X$.
Let $C = \{f_1, f_2, \dots\}$ be a countable dense subset of $X$ (for example, $C$ could be all the functions which are eventually 0) and set $E = X \setminus C$. I claim that $E \notin \mathcal{G}$, but $E$ is a countable intersection of sets from $\mathcal{G}$.
Let $\mathcal{F} = \{ \bigcap_n A_n : A_n \in \mathcal{B}\}$. Note that any set in $\mathcal{F}$ is an intersection of closed sets, hence also closed. (In fact, $\mathcal{F}$ is precisely the set of all closed subsets of $X$.) Suppose, to get a contradiction, that $E \in \mathcal{G}$. Then we could write $E = \bigcup_n F_n$ where $F_n \in \mathcal{F}$ are closed. This means we can write $X = E \cup C = \bigcup_n F_n \cup \bigcup_n \{f_n\}$. By the Baire category theorem, since the singletons $\{f_n\}$ are nowhere dense, one of the sets $F_n$ must be somewhere dense. Since $F_n$ is closed, this means it must have nonempty interior. But that means $F_n$ contains a point of the dense set $C$, which is absurd. So we must have $E \notin \mathcal{G}$. (In particular, since it's not hard to show $E \in \sigma(\mathcal{B})$, we have already contradicted your claim that $\mathcal{G} = \sigma(\mathcal{B})$.)
On the other hand, I claim that for every $f \in X$, we have $X \setminus \{f\} \in \mathcal{G}$. Since $E = \bigcup_n (X \setminus \{f_n\})$, this will show that $E$ is a countable intersection of sets from $\mathcal{G}$. So fix $f \in X$ and let $A_n = \{0,1\}^n \setminus \{f|_{\{1,2,\dots,n\}}\}$, so that $\pi_n^{-1}(A_n)$ consists of all functions $g$ for which there exists $k \le n$ with $f(k) \ne g(k)$. Then $\pi_n^{-1}(A_n) \in \mathcal{B}$ by definition, and we have $X \setminus \{f\} = \bigcup_n \pi_n^{-1}(A_n)$, which shows $X \setminus \{f\} \in \mathcal{G}$.
A similar argument would show $\{f\} \in \mathcal{F}$ for every $f \in X$, implying that $C \in \mathcal{G}$ and showing directly that $\mathcal{G}$ is not closed under complements.
In fact, one should not expect any strategy like this to work. For example, the set of all $\bigcap_k \bigcup_m \bigcap_n A_{nmk}$, where $A_{nmk} \in \mathcal{B}$, does not equal $\sigma(\mathcal{B})$ either. You can pile up any finite number of countable unions, intersections and complements, and you will always be missing some Borel sets. You can even extend this to infinitely many steps, by a transfinite recursion procedure, in which case no countable ordinal number of steps is sufficient. You have to get to the first uncountable ordinal $\omega_1$ to be sure of getting all the Borel sets. This increasing tower of collections of sets is the Borel hierarchy (the collection $\mathcal{G}$ constructed here is $\mathbf{\Sigma}_{2}^0$ in that notation), and it is a theorem that the Borel hierarchy does not stabilize.
