Rudin Theorem 2.41 - Heine-Borel Theorem

When proving Theorem 2.41 in Principles of Mathematical Analysis:

Let $E \subset \mathbb{R}^k$. If every infinite subset of $E$ has a limit point in $E$, then $E$ is closed and bounded.

Rudin says,

"If $E$ is not bounded then $E$ contains points $x_n$ with

$$\vert x_n \vert > n \hspace{2cm} (n = 1,2,3,...).$$

The set $S$ consisting of these points $x_n$ is infinite and clearly has no limit point in $\mathbb{R}^k$, hence has none in $E$."

My first reaction to this was that it is indeed obvious as $S$ is composed of discrete points so I could find a ball around any given point not containing a point of $S$. So, if $p \in \mathbb{R}^k$ is the point I'm constructing a ball around, taking the radius of the ball as $r = \min_{x_n \neq p \in S} \{ \vert x_n - p \vert \}$ should work and as $p$ was arbitrary $S$ doesn't have a limit point. However, I'm not sure if I can take the minimum over an infinite set? I honestly don't know why I shouldn't be able to, but nonetheless I decided to try another proof:

Suppose that $S$ has a limit point $p \in \mathbb{R}^k$. Then for any $r > 0$ $B_r(p)$ has infinitely many points of $S$. So in particular, $B_{\frac{1}{2}}(p)$ contains infinitely many points of $S$. Therefore, there exists a point $y_m \in B_{\frac{1}{2}}(p)$ with $\vert y_m \vert > m > \vert p \vert + 1$ (as otherwise, there would be only finitely many points of $S$ in $B_{\frac{1}{2}}(p)$). Therefore, we have

$$\vert y_m - p \vert \geq \big\vert \vert y_m \vert - \vert p \vert \big\vert > 1 $$

However, this is a contradition as we also have $\vert y_m - p \vert < \frac{1}{2}$. Therefore, as $p$ was arbitrary, $S$ has no limit point in $\mathbb{R}^k$.

My questions: Can I take the min over an infinite set as in my first argument? Is my second argument okay?

I appreciate any answers. Thanks.

Fix an arbitrary point $p\in{\mathbb R}^k$. Put $n_p:=\bigl\lceil|p|\bigr\rceil+1$. All $x_n\in S$ with $n\geq n_p$ satisfy $$|x_n|>n\geq n_p\geq|p|+1\ .$$ Therefore all these $x_n$ are outside the ball of radius $1$ around $p$. It follows that the chosen $p$ cannot be a limit point of $S$, and since $p$ was arbitrary the set $S$ cannot have a limit point in ${\mathbb R}^k$ whatsoever.