Let $x \in \mathbb{R}$ and assume that for all $\epsilon > 0, |x| < \epsilon$. Prove: $x = 0$ 
This following exercise has me kind of confused, it asks: let $x \in \mathbb{R}$ and assume that for all $\epsilon > 0, |x| < \epsilon$.
Prove that $x = 0$.

My attempt to this was to use proof by contradiction:
Proof: Let $x \in \mathbb{R}$ and assume that $x > 0.$ Then our $\epsilon=\dfrac{|x|}{2}>0.$ By assumption we have that $0\le x<\epsilon =\dfrac{ |x|}{2},$ so then $x=0$, which contradicts our $x > 0$ claim.
Will this suffice?
 A: It won't suffice because you have not examined small negative numbers.  
I would prove it by contradiction.  Assume $x \ne 0$.  Then $|x| >0$  Let $\epsilon = |x|/2$.  Then $|x| > \epsilon$, which contradicts the assumption that $|x| < \epsilon$ for every possible $\epsilon > 0$.
A: All of the previous answers invoke contradiction, but I don't believe there's any need to. One could argue like this: By assumption, $|x|$ is smaller than every positive real number, so in particular it is different from every positive real number, so it is not positive. But, by definition, $|x|$ is non-negative. Hence, $|x|$ is zero, so $x$ itself is zero.
A: It's way simpler than that.
$|x|\ge 0$
If $|x|>0$ then setting $\epsilon=|x|$ we get the contradictory $\epsilon =|x| >|x|$.
So $|x|=0$.  So $x=0$.
A: Almost the same proof than E.Fisher, just to use the archimedian property.
$(\mathbb R,+,\le)$ is archimedian, so for $0<|x|<\epsilon$ there exists $n\in\mathbb N$ such that $n|x|>\epsilon$.
This contradicts $|x|<\varepsilon$ for $\displaystyle \varepsilon=\frac{\epsilon}n$, thus $|x|=0\quad$ (and $x=0$ consequently)
