The map $(p,q)\mapsto N(p,q)$ is not necessarily bijection. Take $X=E=\mathbb Z$ as an example. (For any $z\in\mathbb Z$ the ordered pairs $(z,\frac1n)$ are mapped to the same set $N(z,\frac1n)=\{z\}$.)
However, you can take some well-ordering on $E\times\mathbb Q$ and define
$$N(p,q)\mapsto \min\{(a,b)\in E\times\mathbb Q; N(a,b)=N(p,q)\}.$$
(If you prefer, you can get a well-ordering of order type $\omega$ on the set $E\times\mathbb Q$. This can be obtained using any bijection between $E\times\mathbb Q$ and $\omega$. Since you are given enumeration of $E$ and $\mathbb Q$, exhibiting such a bijection does not need axiom of choice.)
The image of this map is a subset of countable well-ordered set $E\times\mathbb Q$. Therefore it is again a countable well-ordered set. So it is in bijection with some countable ordinal.
Additional question in your comment:
How do I show that $G$ is not finite.
If $E$ is infinite, then $G$ cannot be finite.
Just show by induction on $i$ that for each $p_i$ you can find $r_i\in\mathbb Q$ such that $N(p_i,r_i)$ is different from $N(p_j,r_j)$ for each $j<i$.
This can be done by induction, in the inductive step you choose $r_i < \min \{d(p_i,p_j); j<i\}$. (Note that $ \{d(p_i,p_j); j<i\}$ is a finite set of positive real numbers, so the minimum of this set is a positive real number and there exists a rational number smaller than this minimum.)
Then $i\mapsto N(p_i,r_i)$ is an injective map $\omega\to G$. (The point $p_i$ is contained in $N(p_i,r_i)$, but it does not belong to any of the sets $N(p_j,r_j)$ for $j<i$. Therefore any two sets $N(p_i,r_i)$ and $N(p_j,r_j)$, $i\ne j$, are different.)
