Theorem: Let $\{ p_n \} \in X $. If $E \subset X$ and $p$ is a limit point of $E$, then there is a sequence $\{ p_n \}$ such that $p = \lim p_n $
The proof goes like this
proof: For each positive integer $n$, there is a point $p_n \in E$ such that $d(p_n, p) < 1/n$. Given $\epsilon > 0,$ choose $N$ so that $N \epsilon > 1.$ If $n > N$, it follows $d(p_n, p) < \epsilon$. Hence $p_n \to p$
Q1: How come he starts out saying "there is a point $p_n \in E$"? Isn't this what is he trying to prove? The existence of a sequence? He starts out by assuming it exists.
Q2: When he chooses $N$, why does he write it in the following manner "choose $N$ so that $N \epsilon > 1.$"? Why doesn't he write $N > 1/\epsilon$? Is he using Archimedean? How does he know he wants $N \epsilon > 1$? Note that I am not asking "why he chose $N > 1/\epsilon$", I am asking why does he write it like this?
definition:A point $p$ is a limit point of $E \subset X$, if every neighbourhood $N_r(p) \cap E - \{ p \} \neq \emptyset$
Let $n$ be an arbitrary positive integer. Since $\frac{1}{n}>0$, it follows from the definition of a limit point that \begin{equation}B_{\frac{1}{n}}(p) \cap E \setminus \{ p \} \neq \emptyset \, . \end{equation}
Hence there is a point that we denote $p_n$ different from $p$ so that $d(p_n, p)<\frac{1}{n}$. Since $n$ was arbitrary it now follows that, for each positive integer $n$ there is a point $p_n \in E \setminus \{ p \}$ such that $d(p_n,p)<\frac{1}{n}$.
Let $\varepsilon > 0$ be given. By the Archimedean property, there is a positive integer $N$ so that $N\varepsilon>1$. It now follows that $d(p_n,p)<\varepsilon$ whenever $n \geq N$.