Subset of $\ell^2$ with distance property I'd been trying the following problem:

Prove that exists an infinite set $A\subset B(0,1)$ such that $\|x-y\|_2>\sqrt{2}$ for all $x,y \in A$.

My ideas always end in points with distance at most $\sqrt{2}$.
I need hints, please.
 A: Recall that $\ell^2$ is a Hilbert space with the natural inner product. Note that:
$$||x-y||^2=||x||^2+||y||^2-2\langle x,y\rangle,$$
in particular, if $x,y\in S$ ($S$ the set of points with norm equal to $1$)
$$||x-y||^2=2(1-\langle x,y\rangle).$$
So it is enough to find an infinite family $B$ in $\ell^2$ such that their inner product is $\langle x,y\rangle<0$ for all $x,y\in B$, $x\neq y$. Since you can consider
$$A=\left\lbrace x/||x||\;:\;x\in B, x\neq 0\right\rbrace,$$
and we have for $x/||x||$, $y/||y||$, $x,y\in B\setminus\{0\}$ the following
\begin{align} ||x-y||^2 & = 2\left( 1- \frac{1}{||x||\cdot||y||}\langle x,y\rangle\right) \\ & > 2.\end{align}
Let $f:\mathbb{N}\to\mathbb{N}:n\mapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $\ell^2$, and we define 
$$u_n=f(n)e_n-\sum_{k=1}^{n-1}f(k)e_k.$$
The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $\langle u_n,u_m\rangle<0$. In fact
\begin{align}\langle u_n,u_m\rangle & = \left\langle f(n)e_n-\sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-\sum_{j=1}^{m-1}f(j)e_j\right\rangle \\ & = \left\langle f(n)e_n, f(m)e_m\right\rangle-\left\langle f(n)e_n,\sum_{j=1}^{m-1}f(j)e_j\right\rangle \\ & + \left\langle\sum_{k=1}^{n-1}f(k)e_k,\sum_{j=1}^{m-1}f(j)e_j\right\rangle-\left\langle \sum_{k=1}^{n-1}f(k)e_k, f(m)e_m\right\rangle \\ & =  \sum_{k=1}^{n-1}\sum_{j=1}^{m-1}f(k)f(j)\langle e_k,e_j\rangle-f(m)^2 \\ & = \sum_{j=1}^{m-1}f(k)^2-f(m)^2 \\ &\leq mf(m-1)-f(m)^2 \\ & < 0.\end{align}
