$\inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V=0$: Show there must be a sequence $(\vec x^{(n)})$ converging to 0

Question

Let $(V,\rVert \lVert_V)$ be a $k$-dimensional linear space over $\mathbb R$. Let $\Phi:\mathbb R^k\to V$. Choose a norm $\lVert \rVert$ for $\mathbb R^k$. Prove using Bolzano-Weierstrass that $$ \inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V>0. $$

I've chosen the Euclidean norm on the $k$-dimensional Euclidean space $\mathbb R^k$. So we have \begin{align} \lVert\vec x\rVert=\sqrt{x_1^2+\dots+x_k^2}. \end{align}

Assume $\inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V=0$. Consider a sequence $(\vec x^{(n)})$ in $\mathbb R^k$. We know that for each $n\in\mathbb N$, it holds that $\Vert \Phi(\vec x^{(n)})\Vert>0$.

So intuitively, I'd say it's clear that there must be a sequence $(\vec x^{(n)})$ in $\mathbb R^k$, such that $\lim_{n\to\infty} \Phi(\vec x^{(n)})=0$. However, I would like to prove this.

Assume there is no such sequence. That means that for each convergent sequence, the values $\vec x^{(n)}$ will not get arbitrarily close to $\vec 0$. How can I derive a contradiction from here?

EDIT

I think I've got it! We know that for each $n\in\mathbb N$, $\exists \vec x_n:\Vert\Phi(\vec x_n)\rVert<\frac{1}{n}$. So because $\frac{1}{n}\to0$ as $n\to\infty$, it follows that $\lim_{n\to\infty}\Vert\Phi(\vec x_n)\Vert=0.$

Answer 1

This has nothing to do with norms, but only with the definition of the supremum. So you could also try to solve the problem: If $\inf_{x\in A} x = 0$ then there is a sequence $x_n$ in $A$ with $x_n\to 0$. To do so, remember the definition of the infimum.