Proving the pre-image of $[0,1]$ is sequentially compact under this continuous function?

Question

I'm given that to prove that $f:\mathbb R^n\rightarrow \mathbb R$ is continuous and that $\forall u\in \mathbb R^n,$ $f(u)\geq \|u\|.$ I'm then supposed to show that $f^{-1}([0,1])$ is sequentially compact. First of all, we know that $[0,1]$ is sequentially compact. But I'm not sure how to incorporate the fact that $\forall u\in \mathbb R^n,$ $f(u)\geq \|u\|.$ Any suggestions?

Answer 1

If $f(u) \in [0,1]$, then you know that $\|u\| \le 1$. Since $f$ is continuous, you know that $f^{-1}([0,1])$ is closed.

Now use the Bolzano Weierstrass theorem.