Closure and limit of a sequence Let $E$ be a subset of a metric space $(S,d)$. I'm trying to show that an element is in $\overline{E}$ if and only if it is the limit of some sequence of points in $E$.
Suppose there is a sequence $(p_n) \subseteq E$. Then because $E \subseteq \overline{E}$, I know that $(p_n) \subseteq \overline{E}$. Since $\overline{E}$ is closed, $\overline{E}$ contains the limit of every convergent sequence of points in $\overline{E}$. So $p \in \overline{E}.$
This is how I approached the converse: I assumed $p \in \overline{E}$ and that there does not exist a sequence $(p_n) \subseteq E$ such that $p_n \to p$. Then I concluded that there exists  $\varepsilon > 0$ such that $d(p, x) \geq \varepsilon \ \forall x \in E$.
Then there exists an open ball centered at $p$ that does not contain a point of $E$ (I wasn't sure how to make this statement rigorous). So $p \notin \overline{E}$.
Could I get some feedback on my proof?
 A: You haven’t actually justified the conclusion that there is an $\epsilon>0$ such that $d(p,x)\ge\epsilon$ for each $x\in E$. Justifying it properly requires essentially the same work as a direct proof that if $p\in\operatorname{cl}E$, then there is a sequence $\langle p_n:n\in\Bbb Z^+\rangle$ in $E$ converging to $p$, so you might as well give the direct proof.
There are two cases.


*

*If $p\in E$, there’s a trivial sequence in $E$ that converges to $p$; what is it?

*If $p\in(\operatorname{cl}E)\setminus E$, you have to work just a little bit. For each $n\in\Bbb Z^+$ there’s a point $p_n\in E$ such that $d(p,p_n)<\frac1n$; why? What does this get you?
A: Your first part of your proof is fine, as long as you have proved that the limit of a sequence of points in a closed set is part of that set.
There's a bit of a gap in the second part your proof.  Notice that you write:

I assumed $p∈\bar{E}$ and that there does not exist a sequence $(p_n)⊆E$ such that $p_n→p$.

And then conclude:

Then I concluded that there exists $ε>0$ such that $d(p,x)≥ε ∀x∈E$.

But this statement isn't immediately obvious.  What this part requires you to do is actually find a sequence $p_n \to p$.
One way to do this would be to take a sequence of positive real numbers $r_n$ converging to $0$ (like $(\frac{1}{n} )$) and for each real number, find an $x_n \in E$ such that $d(p,x_n) < r_n$ (you know that at least one exists since $\bar{E}$ is closed; $x_n$ might be $p$ itself: it may be useful to divide up cases for when $p \in E$ and when $p$ is a limit point).

 Then for any $\epsilon > 0$, simply find $N$ such that $\forall k \ge N$, $\epsilon > r_k > 0$ so that $d(p,x_k) < r_k < \epsilon$, proving convergence $x_n \to p$.

A: The result is always true in the first countable space. Note that every metric space is first countable.

Let $X$ be first countable and $E$ is a subset of $X$. If $p∈\bar{E}$ iff there  exists a sequence $(p_n)⊆E$ such that $p_n→p$.

