Let $E$ be an unbounded subset of $\mathbb{R}^k$. Then for every $n \in \mathbb{N}$, there exists an $\mathbf{x}_n \in E$ such that $|\mathbf{x}_n| > n$. Let $S$ be the set of all these $\mathbf{x}_n$.
Why is $S$ necessarily infinite? Why does $S$ have no limit point in $\mathbb{R}^k$?
Definition. For any subset $E$ of $\mathbb{R}^k$, a point $\mathbf{p} \in \mathbb{R}^k$ is a "limit point" of $E$ if, for every $r > 0$, there exists some $\mathbf{q} \in E$ such that $0 < |\mathbf{p} – \mathbf{q}| < r$.
Best Answer
If $S$ were finite, $S=\{\mathbf{p}_1,\ldots,\mathbf{p}_k\}$, then letting $M=\max\{|\mathbf{p}_1|,\ldots,|\mathbf{p}_k|\}+1$ you would have that no element of $S$ satisfies $|\mathbf{s}|\gt M$, a contradiction. Thus, $S$ must be infinite.
Likewise, if $\mathbf{p}$ is any point in $\mathbb{R}^k$, then we have that $$|\mathbf{x}_n| = |\mathbf{x}_n-\mathbf{p}+\mathbf{p}| \leq |\mathbf{x}_n-\mathbf{p}|+|\mathbf{p}|.$$ In particular, $|\mathbf{x}_n-\mathbf{p}|\geq n-|\mathbf{p}|$ for all $n$.
Let $N$ be the smallest positive integer such that $N\gt|\mathbf{p}|$. Then for $n\geq N+1$ we have that $|\mathbf{x}_n-\mathbf{p}|\geq n-|\mathbf{p}|\geq N-|\mathbf{p}|+1\gt 1$.
Now let $$r = \frac{1}{2}\min\Bigm\{|\mathbf{x}_1-\mathbf{p}|,\ldots,|\mathbf{x}_N-\mathbf{p}|, 1\Bigr\}.$$ If $r\gt 0$ (that is, $\mathbf{p}\neq \mathbf{x}_n$ for all $n$), then for every $n\in\mathbb{N}$ we have $|\mathbf{x}_n-\mathbf{p}|\gt r$ (if $n\leq N$, then by construction we have $r\leq \frac{1}{2}|\mathbf{x}_n-\mathbf{p}|\lt|\mathbf{x}_n-\mathbf{p}|$, and if $n\geq N+1$, then $|\mathbf{x}_n-\mathbf{p}|\gt 1 \gt r$).
If $r=0$, then $\mathbf{p}=\mathbf{x}_i$ for some $i\leq N$. Replace $r$ with $$\frac{1}{2}\min\Bigl(\Bigl\{ |\mathbf{x}_m-\mathbf{p}|\;\Bigm|\; \mathbf{x}_m\neq\mathbf{p}, m\leq N\Bigr\}\cup\{1\}\Bigr)$$ and proceed as above.