hint needed: Identity mapping from ($K^n, ||.||_2) \to (K^n, ||.||)$ is continuous, where ||.|| is arbitrary

Question

I am thinking of saying by Munkres Theorem 18.1 part 4, for all x and $\epsilon$ > 0, $\exists \delta > 0 $ s.t. x $\in B_2(x,\delta) \subset B(f(x),\epsilon)$ Then trying to show this not possible using properties of $||.||_2$ metric. But am not sure this is the right approach or not.

Answer 1

Hint: prove that the identity is continuous iff the two norms are equivalent. What do you know about norms on a finite dimensional vector space?