As I understand it, he is defining the set $\{q_1, q_2, ... , q_n\}$ as the image of a bijective function whose domain $D$ is the intersection of $N$ and $E$. My question is: How we can verify that this image can exist as a subset of the same metric space $X$ (with $p$, $N$, and $E$) so that $d(p,q_i)$ is even defined? 
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The definition of limit point is as follows:
$$\text{$p\in E$ such that for all $\epsilon>0$, there is $q\in E\backslash\{p\}$ such that $d(p,q)<\epsilon$}$$
Let $X$ be a metric space. Suppose $p$ is a limit point of $E\subseteq X$. Suppose that for some $r>0$ there is $N=\{x\in E: d(p,x)<r\}$ such that $(N\cap E)\backslash\{p\}=\{q_1,q_2,...,q_n\}=A$.
$A$ is simply the intersection between two sets with one element (namely $p$) removed; there are no functions involved. If you wish, you can define $f:A\to A$ such that $f(x)=x$, but this really isn't necessary.
$X$ is a metric space, so the metric $d$ on $X$ exists by definition. Since $A \subseteq X$ and $p\in E \subseteq X$, $d(p,q_i)$ is defined for all $q_i\in A$.