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!
 A: Two answers have already pointed out the theorem that all norms on $\mathbb R^n$ induce the same topology. So they agree as to which sets are open, which are closed, which are compact, etc. But I don't think that completely answers the question, because different norms have different closed unit balls.  (To avoid ambiguity in the notation $\Vert x\Vert$, I'll use $E$ for the standard Euclidean norm and $N$ for some arbitrary other norm.) We need to show that the unit $N$-ball $B=\{x:N(x)\leq1\}$ is compact, in the topology induced by $N$ --- or by $E$ or by any other norm, since they're all the same topology. 
If (and it's a big "if") you already know that closed bounded sets are compact, for the arbitrary norm $N$, then this is easy.  $B$ is obviously bounded with respect to $N$.  It's closed because the norm function $N$ is continuous, thanks to the triangle inequality which implies $|N(x)-N(y)|\leq N(x-y)$.  So $B$ is compact.
But what if you know "closed bounded sets are compact" only for the standard Euclidean norm $E$? Then I want to prove that $B$ is closed and bounded with respect to $E$. The "closed" part of this is easy, since all norms agree as to which sets are closed, and I pointed out in the previous paragraph that $B$ is closed with respect to $N$.  So it remains to check that $B$ is bounded with respect to $E$.
I'll show this by contradiction. So suppose $B$ is not bounded with respect to $E$, i.e., there is, for each positive integer $n$, some $x_n\in B$ with $E(x_n)\geq n$.  Consider the sequence of vectors $(x_n/n)$.  Each of these vectors has $E$-norm $\geq1$, so the sequence doesn't converge to $0$ with respect to $E$. But, since all the $x_n$ are in $B$, we have $N(x_n/n)\leq1/n$, so the sequence does converge to $0$ with respect to $N$.  This contradicts the fact that $E$ and $N$ induce the same topology.
Confession: In the last paragraph, I proved a claim about boundedness using the fact that all norms induce the same topology.  As far as I'm aware, any proof that all norms induce the same topology will include information from which one can much more easily derive the claim about boundedness.  That information is likely to be of the form "there exists a constant $M$ such that, for all $x$, we have $E(x)\leq M\cdot N(x)$."  My excuse for giving the argument in the form I used is that the two previous answers and the earlier part of my answer all involved the fact that norms on $\mathbb R^n$ induce the same topology, so it seemed efficient to keep on using the same fact.
EDIT: To save people the bother of checking the timing of answers, the two previous answers I mentioned above are those of Theo and Yaddle.  Eric Wofsey's answer arrived while I was typing mine, and it includes the additional information that I said was lacking in the previous answers.
A: This follows from the fact that all norms on $\mathbb{R}^n$ are equivalent.  In particular, let $\|\cdot\|$ be your norm and let $\|\cdot\|_2$ be the Euclidean norm.  Since the norms are equivalent, they generate the same topology, so the closed unit $\|\cdot\|$-ball is also closed in the Euclidean topology.  Moreover, the norms being equivalent implies there is a constant $K$ such that $\|x\|_2\leq K\|x\|$ for all $x$.  This means that the closed unit $\|\cdot\|$-ball is contained in the closed Euclidean ball of radius $K$.  Thus the closed unit $\|\cdot\|$-ball is closed and bounded with respect to the Euclidean topology, and hence compact.  (It is compact with respect to either norm, since the two topologies are the same.)
A: The closed unit ball is compact in $\mathbb R^n$ respecting every norm. That is due to the fact that all norms on $\mathbb R^n$ are equivalent (and hence generate the same topology). 
In general you can show that on every finite dimensional Banach space all norms are equivalent and hence the unit ball is compact respecting every norm. 
You can even show that a Banach space is from finite dimension if and only if the unit sphere is compact. That is sometimes pretty handy in functional analysis. I hope that it helps you :) 
A: 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. 
