Let $E$ be a subset of $R^{k}$, with the Euclidean metric on it. Then $1.$ implies $2.$, where
$1.$ Every infinite subset of $E$ has a limit point in $E$
$2.$$E$ is closed.
Proof Suppose by contradiction that $E$ is not closed. Since by definition a subset of a metric space is said to be closed if it contains all its limit points, we are assuming there exists a point $x_0$ of $\mathbb{R}^k$ which is a limit point of $E$ and which lies in the complement of $E$ with respect to $\mathbb{R}^k$. For every $n\in\mathbb{N}$, there are points $x_n\in E$ such that $|x_n-x_0|<\frac{1}{n}$. Let $S$ be the set of these points. Then $S$ is infinite (otherwise $|x_n-x_0|$ would have a constant positive value, for infinitely many $n$)....
I can't understand the "otherwise" argument. Since He said $S$ is infinite, I suppose that "otherwise" means that $S$ is by contradiction assumed to be finite. But then how can we consider infinitely many indices $n$, so that $x_n$ is a point of $S$ for each $n\in\mathbb{N}$? I think the only way is to "count" some of them infinitely many times. But then how can the condition $|x_n-x_0|<\frac{1}{n}$ hold for all of them, considering that $\frac{1}{n}$ converges to $0$, the only point $x_n$ with $|x_n-x_0|=0$ is $x_0$ and $x_0\notin S$?
Could you please expand Rudin's argument so to make it more understandable to me? Thank you.