Constructing non-Cauchy sequences in an infinite-dimensional space Some proof I stumbled on uses that

In an infinite-dimensional space we can find a sequence $(x_n)_{n\geq 1}$ such 
  that $\Vert x_n\Vert \leq 1$ and $\Vert x_n-x_m\Vert\geq 1/2$ for all $n\neq m$.

I am new to infinite-dimensional spaces. Could someone explain why this would be true?
 A: That's a corollary of a well-known lemma of Riesz:

Lemma (Riesz). Let $X$ be a normed vector space, $U \subsetneq X$ a proper closed subspace, and $\delta > 0$. Then there is $x \in X$ with $\def\norm#1{\left\|#1\right\|}\norm{x} = 1$ such that 
  $$ \inf_{u \in U} \norm{x-u} > 1- \delta. $$

For your statement let $\delta = \frac 12$. By induction define a sequence $(x_n)$. Choose $x_1$ with $\|x_1\| = 1$ arbitrary. If $x_i$ for $i< n$ are chosen, let $U_n := \mathop{\rm span}\{x_i: i < n\}$. Then $U_n$ is finite-dimensional, hence a proper closed subspace of the infinite-dimensional space $X$. By Riesz, there is $x_n \in X$ with $\norm{x_n} = 1$ and 
$$ \inf_{u \in U_n} \norm{x_n-u} > \frac 12. $$
As $x_i \in U_n$ for $i < n$, we have 
$$ \norm{x_i -x_n} > \frac 12, \qquad i < n $$
A: Let us consider for example the space $\ell_\infty$ of all bounded complex sequences, equipped with the norm defined by :
$$\Vert u\Vert_\infty=\sup_{n\in\mathbb{N}}\vert u_n\vert$$
If $u^{(r)}$ denotes the sequence $(\delta_{n,r})_{n\in\mathbb{N}}$ (where $\delta_{i,j}$ is the Kronecker symbol), we can readily see that :
$\forall (p,q)\in\mathbb{N}^2,\,p\neq q\Rightarrow\Vert u^{(p)}-u^{(q)}\Vert_\infty=1$
This example should help understanding the general case ...
