Will the "closed" unit ball $\left\| x \right\| \le 1$ in $\Bbb R^n$ be a compact set for any norm? 
Will the "closed" unit ball $\left\| x \right\| \le 1$ in $\Bbb R^n$ always be a compact set for any norm? 

I am asking this question because the induced matrix norm is originally defined as a supremum,

Given a vector norm ${\upsilon}:\Bbb R^n\to \Bbb R$, the induced matrix norm $\mu:\Bbb R^{m\times n} \to \Bbb R$ is defined as $\mu({\bf{A}})=\sup\{{\bf{Ax}}:\upsilon({\bf{x}})\le 1\}$.

However, my lecture material later claims the closed ball "$\upsilon({\bf{x}})\le 1$" is compact so the supremum is actually a maximum, so



I don't know where this lector material comes from and I have googled but cannot find the Corollary 2.153. Hope someone can help with this. Thank you!

Yes,  any two norms are equivalent on a finite dimensional Banach space, so they generate the same topology. 
Perhaps for more detail, it is fairly easy to show that the unit ball under the $l_1$ norm, $B(l_1^n)$, is compact. If a sequence is bounded in the $l_1$-norm then it bounded coordinate-wise, and now apply Bolzano-Weierstarss (to the first coordinates, then to the second, and so on)  to select a sequence that converges point-wise. In $l_1$ is then immediate that pointwise convergence of the sequence implies norm convergence. 
Since any two norms are equivalent,  $B(l_1^n)$ is a compact set in the topology determined by other norm $||\cdot||$ and the unit ball $B_{||\cdot||}$ is a closed subset of a multiple of $B(l_1^n)$, thus also compact. 
