If $|x|<\epsilon, \forall \epsilon > 0$, then $x=0$

Suppose $x\neq 0$. Then $|x|>0$. Take $\epsilon = |x|/2$ and we have $|x|<\epsilon = |x|/2$, a contradiction.

First of all let's say you are wrong, $x\ne 0$, now if $x\ne 0$ then $|x|>0$. Because you said for all $\epsilon$ any expression of $epsilon$ that is greater than $1$ should be valid. So let's try $\epsilon\equiv \frac{|x|}{n}, n\in\Bbb R^+, n>1$, because both $n$ and $|x|$ are positive $\epsilon$ is also positive. Now it is given that $|x|<\epsilon$ so $$|x|<\epsilon\implies |x|<\frac{|x|}{n}\implies 1<\frac{1}{n}$$because that $n>1$ we have contradiction. So $x=0$