Is there other way to prove that unit sphere in Euclidean space is a infinite set? Let $S=\{x\in \mathbb{R}^n;|x|=1\}$ the unit sphere in Euclidean space, where $n>1$ and $|\cdot|:\mathbb{R}^n\rightarrow\mathbb{R}$ is a arbitrary norm.
The problem is to prove that $S$ is a infinit set. One way to prove it, is to show that there is a infinit set $X\subset \mathbb{R}^n$ such as $|x|=1$ for all $x\in X$. For example, $$X=\left \{\frac{(m,1,...,1)}{|(m,1,...,1)|}\in\mathbb{R}^n;m\in\mathbb{N}\right\}$$
Is there other way?
 A: Try the 2-dimensional case and embed inductively circle of inferior dimension in "bigger" ones.
A: First, try to do it for $\|\cdot\|_2$:
$$f:\begin{array}{ll}[0,1]\to S\\t\mapsto \left(t,\sqrt{1-t^2},0,\dots,0\right)\end{array}$$  
$f$ is injective since $\forall t_1,t_2 \in [0,1], f(t_1)=f(t_2) \implies \left(t_1,\sqrt{1-t_1^2},0,\dots,0\right) = \left(t_2,\sqrt{1-t_2^2},0,\dots,0\right) \implies t_1 = t_2$. This tells you that there are more elements in $S$ than in $[0,1]$. Since you know that $[0,1]$ is infinite, youcan therefore conclude that $S$ is.

But for another norm, this can't work since our function was made specifically for $\|\cdot\|_2$.
Let's start by taking $\alpha = \cfrac{1}{\|(1,0,\dots,0)\|}$. We have $\|(\alpha,0,\dots,0)\|=1$.
Now for any $x\in [0,\alpha],$ define
$$g_x:\begin{array}{ll}\Bbb R_+\to \Bbb R_+\\t\mapsto \left\|\left(x,t,0,\dots,0\right)\right\|\end{array}$$
$g_x(0)=\cfrac{x}{\alpha}\le 1$ and $\lim\limits_{t\to +\infty}g_x(t)=+\infty$ so by the mean value theorem, we have $t_0 \in \Bbb R_+,g_x(t_0)=1$. For any $x\in[0,\alpha]$, we'll call that $t_0$ $h(x)$.
Now define $$f:\begin{array}{ll}[0,1]\to S\\t\mapsto \left(t,h(t),0,\dots,0\right)\end{array}$$
It is properly defined since we chose the $h(t)$ so that the vector is in $S$ and it is still injective for the same reason as for $\|\cdot\|_2$. So $S$ is infinite.
