For any $\epsilon$, if $\epsilon>0$ and $|x|<\epsilon$, then $x=0$ For any $\epsilon$, if $\epsilon>0$ and $|x|<\epsilon$, then $x=0$.
I understand that supposing  $\epsilon=\frac{x}{2}$ will lead to a contradiction, but let’s take a correct case:
Let $\epsilon=3$, then $x$ would have a whole set of values. Can you explain what is going on? 

If it is true for any real $\epsilon>0$, it will be true in particular, for any rational $r>0$,
so we will have, for any natural $n $,
$$|x|<\frac {1}{n+1} $$
or
$$-\frac {1}{n+1}<x <\frac {1}{n+1} $$
and the squeeze theorem gives
$$0\le x\le 0$$
The downvoter is a zam.
