How to choose sequence in not totally bounded metric space whose distance remain greater than positive? Suppose that $(X, d)$
 be a metric space. If $\epsilon  >0$ is given, a subset $A\subset X$ is said to be an $\epsilon-$net if $A$ is finite and $X= \cup_{a\in A}S_{\epsilon}(a),$ where $S_{\epsilon}(a)=\{x\in X: d(x, a)< \epsilon \}.$ A metric space $X$ is said to be totally bounded if it has an $\epsilon-$net for every $\epsilon>0.$
Assume that $X$ is not a totally bounded. So, for every finite subset $A\subset X$, there exists $\epsilon_{1}>0$ such that  $X\neq \cup_{a\in A} S_{\epsilon_{1}}(a).$  In this case, we note that  $\cup_{a\in A} S_{\epsilon_{1}}(a) \subsetneq X$, that is, there exists $x\in X$ such that $d(x, a) > \epsilon_{1}$ for all $a\in A.$

My Question: Assume $X$ is not a totally bounded. Can we say there is a sequence $\{x_{n}\} \subset X $  and $\delta >0$ such that $d(x_{n}, x_{n'})\geq \delta$ for $n\neq n'$?


Yes, but you do have to prove it. You can prove it by recursively constructing such a sequence. If $X$ is not totally bounded, there is an $\epsilon>0$ such that whenever $F\subseteq X$ is finite, 
$$\bigcup_{x\in F}B(x,\epsilon)\ne X\;.$$
Let $x_0\in X$ be arbitrary. Then $B(x_0,\epsilon)\ne X$, so let $x_1\in X\setminus B(x_0,\epsilon)$. In general, given $x_0,\ldots,x_n$, we know that
$$\bigcup_{k\le n}B(x_k,\epsilon)\ne X\;,$$
so we can pick $x_{n+1}\in X\setminus\bigcup_{k\le n}B(x_k,\epsilon)$. It’s now straightforward to verify that $d(x_m,x_n)\ge\epsilon$ whenever $m,n\in\Bbb N$ and $m\ne n$.
