Given $X$ metric space and $E$ is any subset of $X$.
If $x$ is a limit point of $E$, for every $\epsilon>0$, prove that neighborhood of x contains infinitely many element
I use the fact that when we have $x$ as a limit point of $E$, we can find a sequence on $E$ that converges to $x$, and because the sequence on $E$ and subset of neighborhood of $x$, we can conclude that neighborhood of $x$ contain infinitely many elements
Does my argument make sense? I hope you can tell me if I'm wrong guys
Let $V$ be a neighborhood of $x$. Then there is a $r>0$ such that $B_r(x)\subset V$. Take $N\in\mathbb N$ such that $\frac1N<r$. For each $n\in\mathbb N$, take $x_n\in B_{\frac1{n+N}}(x)\setminus\{x\}$. Then the set $S=\{x_n\,|\,n\in\mathbb N\}$ must be an infinite set; otherwise, let $\varepsilon$ be the smallest distance from $x$ to an element of $S$. Then we would always have $d(x,x_n)\geqslant\varepsilon$, which is impossible, since $d(x,x_n)<\frac1{n+N}$. So, $S$ is an infinite set and $S\subset V$.
In your proof, you did not prove that your sequence has infinitely many distinct elements.