Definition
If $V$ is a finite dimensional vector space then we say that the norm $||\cdot||_1$ and $||\cdot||_2$ are equivalent if and only if there exist two positive constant $m$ and $M$ such that $$ m||v||_1\le||v||_2\le M||v||_2 $$ for any $v\in V$.
Lemma
If $V$ is a finite dimensional vector space then the function $||\cdot||_\infty:V\rightarrow\Bbb R^+_0$ defined through the condition $$ ||v||_\infty:=\max_{i=1,...,n}|v_i| $$ is a norm in $V$.
Theorem
If $V$ is a finite dimensional vector space then all norm in $V$ are equivalent.
Clearly if $||\cdot||$ is a norm in $V$ then by triangle inequality $$ ||v||\le M\cdot||v||_\infty $$ where $M:=\max_{i=1,...,n}||e_i||$ for any basis $\mathcal{B}:=\{e_1,...,e_n\}$.
Unfortunately I can't prove the other inequality. So could someone help me, please?
Method 1
Suppose There are no $D>0$ s.t. for all $x\in V$, $\|x\|\geq D\|x\|_\infty $. This mean that $$\forall n\in\mathbb N^*, \exists x_n\in V: \|x_n\|\leq \frac{\|x_n\|_\infty }{n}.$$
Instead of considering $y_n=\frac{x_n}{\|x_n\|_\infty }$, we can suppose WLOG that $\|x_n\|_\infty =1$ for all $n$. Therefore, by Bolzano-Weierstrass, there is a subsequence (still denoted $(x_n)$) that converges to $x$. Since a norm is continuous, you have $$\|x\|=\lim_{n\to \infty }\|x_n\|=0,$$ and thus $x=0$, but on the other hand, $$\|x\|_\infty =\lim_{n\to \infty }\|x_n\|_\infty =1,$$ which is a contradiction.
Method 2
You have that $\|\cdot \|$ is continuous on $$S=\{x\in V\mid \|x\|_\infty =1\},$$ which is compact. Therefore, there are $x_0,x_1\in S$, s.t. for all $x\in S$, $$0<\|x_0\|\leq \|x\|\leq \|x_1\|.$$ Since $$S=\left\{\frac{y}{\|y\|_\infty }\mid y\in V-\{0\}\right\},$$ you get the wished result.