Relationship between general norm and infinite norm in $\mathbb{R}^k$ We work in the vector space $\mathbb{R}^k$ over $\mathbb{R}$. We define the (usual) infinite norm  on $\mathbb{R}$:
$$\|x\|_{\infty} = \max\{|x_1|,|x_2|,\dots,|x_n|\}$$ where $x_1, x_2, \dots, x_n $ are the coordinates of of $x$ in $\mathbb{R}^k$. 
We would like to prove the following fact:
$$\exists c> 0, \forall x\in \mathbb{R}^k, \|x\|_{\infty} \le c \|x\|$$
where $\|\cdot \|$ is any norm. I am also told that if this isn't true, then I can construct a sequence $(x_n)$ such that $\|x_n\|$ is bounded but $\|x\|_{\infty}$ diverges to infinity.
I am slightly puzzled by that hint, and I'm stuck trying to prove the existence of such a sequence. Here is what I've done: first negate the statement:
$$\forall c> 0, \exists x\in \mathbb{R}^k, \|x\|_{\infty} > c \|x\|$$
Then I thought, whynot construct the obvious sequence and see where that takes us? So we define $(x_n)$ such that $$\|x_n\|_{\infty} > n \|x_n\|$$ This at least tells us that $\frac{\|x_n\|}{\|x_n\|_{\infty}}$ tends to $0$, but I haven't gotten much further. What bugs me the most is trying to somehow bound $\|x_n\|$. 
Could someone suggest an approach to tackle that hint?
 A: Suppoe no such $c$ exists. Then for every $r>0$ there exists $x_r$ such that $\|x_r\|_{\infty}>r\|x\|.$ 
So for $n\in \Bbb N$ let $\|x_{n^2}\|_{\infty}>n^2\|x_{n^2}\|.$ This requires $\|x_{n^2}\| \ne 0.$ 
Let $y_n=\frac {x_{n^2}}{n\|x_{n^2}\|}.$ $$\text { Then }\quad \|y_n\|=\frac {1}{n}$$ $$\text { and }\quad  \|y_n\|_{\infty}=\frac {\|x_{n^2}\|_{\infty}}{n\|x_{n^2}\|}> \frac {n^2\|x_{n^2}\|}{n\|x_{n^2}\|}=n.$$
BTW we can also observe that for $x=(x_1,...,x_k)$ there exists $j$ with $1\leq j\leq k$ and $|x_j|=\|x\|_{\infty}$ so $|x_i|\leq |x_j|$ for $i=1,...,k .\quad$  Therefore $$\|x\|=\left(\sum_{i=1}^kx_i^2\right)^{1/2}\geq (x_j^2)^{1/2}=|x_j|=\|x\|_{\infty}$$ and  also  $$\|x\|=\left(\sum_{i=1}^kx_i^2\right)^{1/2} \leq \left(\sum_{i=1}^kx_j^2\right)^{1/2}=(kx_j^2)^{1/2}=\sqrt k\;|x_j|=\sqrt k\; \|x\|_{\infty}.$$ 
A: You can choose the sequence $(x_n)$ such that $\|x_n\|_\infty=1$. Then $\|x_n\|\to0$.
To arrive at a contradiction, I will use the Weierstrass theorem. First, we show that $x\mapsto \|x\|$ is continuous wrt the $\infty$-norm:
Write $x=\sum x_ie_i$ with the unit vectors $e_i$, then:
$$
\|x\|\le \sum_{i=1}^n |x_i| \|e_i\| \le \|x\|_\infty \sum_{i=1}^n \|e_i\|.
$$
Then for arbitrary $x,y$, we get
$$
|\|x\|-\|y\||\le \|x-y\|\le \|x-y\|_\infty \sum_{i=1}^n \|e_i\|.
$$
Then the continuous function $x\mapsto \|x\|$ attains its minimum on $\{x:\|x\|_\infty=1\}$. Since $\|x\|\ne0$ on this set, there is $c>0$ such that $\|x\|\ge c$ for all $\|x\|_\infty=1$. 
Now, going back to the original sequence, we arrive at a contradiction.
