Is $0$ the only number which satisfies $-\epsilon 0$? The question I am trying to answer is:
Can you find more than one number $a$ which satisfies
(a)$-\epsilon <a< \epsilon$, $\forall \epsilon >0$;
(b)$|a|<\epsilon$, $\forall \epsilon>0$ (Argue by contradiction. If $a\neq 0$, try $\epsilon=\frac{1}{2}|a|$).
Source: Numbers and Functions by R. P. Burn. Chapter:2 Question: 65

Working:
(a) Since $a$ has to be less than every positive $\epsilon$, the only number which can satisfy this condition is $0$. Taking any other number would result in some value of $a$ not satisfying the inequality at $a=\epsilon$.
Edit (a): By choosing an arbitrarily 'small' $\epsilon$ we get $-\epsilon$ getting arbitrarily close to $a$ from the negative side and $\epsilon$ getting arbitrarily close to $a$ from the positive side. It makes intuitive sense then that the only number for which $-\epsilon<a<\epsilon$ holds true is $a=0$.
(b) $-\epsilon <a<\epsilon \Leftrightarrow |a|<\epsilon$, so the same holds $\forall \epsilon >0$.
(Contradiction argument: for $\epsilon=\frac{1}{2}|a|$ and $a\neq 0$, we would have $|a|<\frac{1}{2}|a| \Rightarrow 1<\frac{1}{2}$, which is not true.)
Doesn't the above also suggest that when $|a-b|<\epsilon, \forall \epsilon >0$, $a=b$?
 A: If $a$ has to be a real number, then that is correct.
But just for fun, would this hold in other models of the usual axioms for the real numbers?
Turns out that for other fun models such as non-standard analysis the answer is still YES.
Indeed, the usual axiomatizations of the reals such as Tarski's constrain their models to be linearly ordered fields, where every number other than zero is either positive or has a positive opposite.
Since the hypothesis requires than $-\epsilon < a < \epsilon$ for every $\epsilon$, and in particular $a < b$ implies $a \neq b$, we have that only the element zero satisfies this property.
A: For the part (b), just notice the equivalence
$$-\epsilon<a<\epsilon\iff|a|<\epsilon.$$
To address the initial uniquenes question directly, let us assume that there are two numbers $a\ne b$ such that 
$$\forall \epsilon>0,-\epsilon<a,b<\epsilon.$$
But by hypothesis $\eta:=|a-b|$ is a positive number, and $|a-b|<|a|+|b|<2\epsilon$, so that all $\epsilon$ such that $0<\epsilon\le\eta/2$ create a contradiction.
