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? 
 A: You misunderstand the role of "for any" there. Stated correctly, this is an assertion across all possible values of $\epsilon$, and as such: if it is always true that $|x|<\epsilon$ (no matter which $\epsilon $ is chosen) then we draw the conclusion that $x=0$
$x:(\forall \epsilon>0: |x|<\epsilon)\implies x=0$
A: As you have stated it,
For any $\epsilon$,  if $|x| <\epsilon$, then $x=0$, 
the assertion is false and your counter-example shows that.
A true statement that may sound similar is:
If for every $\epsilon > 0$, $|x| <\epsilon$, then $x=0$.
This is a different statement that the first one.   This second statement says that if it is true that $|x|$ is less than every positive number, the $x=0$.  And that statement is true.
A: 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.
