The resource of the proof is : http://mypage.concordia.ca/mathstat/pgora/m364/Density.pdf
The proof used the pigeon-hole principle for infinite cases which was strange.
The proof is:
Denote by $<x>=x-\lfloor x \rfloor$ the fractional part of the real number $x$, which is a number in $[0,1)$. Then I claim that
$\forall \epsilon >0$, $ \exists n_{0},m_{0} \in \mathbb{N} $: $|n_{0}-2 \pi m_{0}|< \epsilon$.
Once this is proved, the lemma also will follow from the properties of inf. To prove the claim consider all the numbers $<2 \pi a>, a \in \mathbb{N}:$they all belong to $[0,1)$ and since they are infinitely many, by the Pigeonhole principle there are two (distinct!) $a,b \in \mathbb{N}$ which are less than $\epsilon$ apart from each other:
$\epsilon > |<2 \pi a>-<2 \pi b>|=|2 \pi (a-b)- \lfloor 2 \pi a \rfloor + \lfloor 2 \pi b \rfloor|$
We can assume $a<b$ (if not, just rename the numbers) and hence we have the claim with $n_{0}=\lfloor 2 \pi a \rfloor - \lfloor 2 \pi b \rfloor$ and $m=b-a$.
Let $k\in\mathbb{N}$. Take any $k$ positive integers, say $a_1,\dots, a_k$. In order for the result to be false, all the open balls $B_{\epsilon/2}(<2\pi a_i>)$ must be pairwise disjoint. However, since $k$ is arbitrary, and all $<2\pi a_i>$ must lie within $[0,1)$, we can choose $k$ large enough so that there must be some overlap between some distinct open balls. This is where pigeonhole principle is used. I'll leave the formal details to you.