Let $X$ be a metric space. Let $ E \subset X$ and let $p$ be a limit point of $E$. Prove that there is a sequence $ \{p_n\} $ in $E$ such that $p= \lim_{n\to\infty}p_n$.
This theorem is from page 48 of Rudin's Principles of Mathematical Analysis. I've tried to prove the statement. Is this approach okay?:
Suppose that there is no sequence $\{p_n\}$ in E such that $p=\lim_{n\to\infty}p_n$.
Then $\exists \epsilon >0$ such that $\forall N$, $n \geq N$ implies that $d(p,p_n) \geq \epsilon$.
So $p_n \notin V_\epsilon (p)$ as $p_n \rightarrow p (n \geq N)$, where $V_\epsilon (p)$ is neighbourhood of point $p$ with radius $\epsilon$.
Then it is not true that all nbd's of p contain all but finitely many terms of $\{p_n \}$, because for $\epsilon$ above, the nbd of p, $V_\epsilon (p)$, does not contain infinitely many points of $p$.
Hence $p$ is not a limit point of E.
Therefore, by contraposition, if $E \subset X$, and $p$ is limit point of E, then
there exists a sequence $ \{p_n \}$ in E such that $p= \lim_{n\to\infty} p_n$.
Best Answer
The way you said that there is no sequence of $E$ that converge to $p$ is wrong. To say that there is no sequence of $E$ that converge to $p$ is $$\exists \varepsilon>0: B(p,\varepsilon)\cap E=\emptyset,$$ where $B(x,r)=\{y\in X \mid d(x,y)<r\}$ (here, we let $d$ denote the metric of the metric space $X$).
Notice that your approach is weird (and unclear). If $p$ is a limit point of $E$, (I denote $E_p:=E\backslash \{p\}$) then $$\forall \varepsilon>0, B(p,\varepsilon)\cap E_p\neq \emptyset,$$ in other words, $$\forall n\in\mathbb N^*, \exists x_n\in E_p\cap B\left(p,\frac{1}{n}\right),$$ therefore $(x_n)$ is a sequence of $E$ that converges to $p$.