Let $S$ be a non-empty subset of $\mathbb{R}$and $a\in\mathbb{R}$. Prove that $a\in\overline{S}$ if and only if for each positive integer $n$,there exists an $x_n\in S$ such that $|x_n-a|<\frac{1}{n}$.
If there was $(x_n)_{n\in\mathbb{N}}$ such that $|x_n-a|>\frac{1}{n}\forall n$ Then for $n>\mathbb{N}$and for$\epsilon>0$ $$|x_n-a|>\epsilon$$, then there exists the following open intervals $\mathscr{U}=(a-\frac{\epsilon}{2},a+\frac{\epsilon}{2})$ and $B=(x_n-\delta,x_n+\delta)$ for $0<\delta<\frac{\epsilon}{2}$ such that $\mathscr{U}\cap B=\emptyset$ then $a\notin A'$ and $a \notin A$ which concludes $a\notin \overline{A}$.
So if $a\in \overline{A}$ then there must be a sequence that $x_n\in S$ such that $|x_n-a|<\frac{1}{n}$.
Question:
Is my proof right? If not. How should I prove the claim?
$"\Leftarrow"$ Take any $\epsilon>0$. (We neeed to see $S\cap (a-\epsilon,a+\epsilon)\neq \emptyset$.) Then there must be $n$ such that $\frac{1}{n}<\epsilon$ and $|x_n-a|<\frac{1}{n}<\epsilon$. So $x_n \in S\cap (a-\epsilon,a+\epsilon)$.
$"\Rightarrow"$ $a\in \overline{S}$ means that $S\cap (a-\epsilon,a+\epsilon)\neq \emptyset$ for any $\epsilon>0$. For every $n>0$ say $\epsilon_n=\frac{1}{n}$, and choose some $x_n \in S\cap (a-\epsilon_n,a+\epsilon_n)$. Then we get a sequence $(x_n)$ and it has property what we want.
Addition: I am sorry your attempt can not work. For ex: $S=(0,3)$, $a=3$ and $x_n=1$ for all $n$. If you want to prove with contradiction you need to see for any sequence $(x_n)$ has not that property.