This comes from an exercise in Appendix C from Axler's Measure, Integration & Real Analysis. The following is my approach.
Suppose $b \neq 0$. Let $|b| = \epsilon$. Then by Archimedean Property (2) $$\exists n^* \in \mathbb{Z}^+ \text{ such that} \frac{1}{n^*} < \epsilon$$ but $b < \frac{1}{n}$, $\forall n \in \mathbb{Z}^+$. Hence a contradiction.
Am I approaching this correctly, if I am are there any other approaches?
You are approaching it correctly and your approach is the simplest and intuitive way.
But if you wish to be direct and analytical you could do the following.
$b\in \mathbb R$ so there exist a sequence of $q_n \in \mathbb Q$ so that $q_n \to b$.
For any $\epsilon > 0$ there exist an $N$ so that $n > N$ implies $|q_n - b| < \frac {\epsilon}2$. But there is also a $k\in \mathbb N$ so that $0 < \frac 1k < \frac \epsilon 2$.
So $|q_n- 0| \le |q_n - b| + |b-0|$. $|q_n - b| < \frac \epsilon 2$ and $|b-0| = |b| < \frac 1k < \frac \epsilon 2$ so $|q_n - 0| < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$.
So $\lim_{n\to \infty} q_n = 0$. But $b = \lim_{n\to \infty} q_n $. So $b = 0$.
But... I'd prefer to read your proof than my proof.
Assume that $a \neq 0.$ Then $|a| \neq 0.$ Take $N=\lfloor\dfrac{1}{|a|}\rfloor+1 \in \mathbb{Z_+},$ where $\lfloor x \rfloor$ denotes the floor integer function. Then $$\frac{1}{|a|}<\lfloor\dfrac{1}{|a|}\rfloor+1=N,$$ which implies that $$|a|>\frac{1}{N},$$which contradicts.
Your approach is correct. There are several ways to define $\Bbb R$ from $\Bbb Q.$ They all result in isomorphic structures... Up to isomorphism there is only one ordered field in which every non-empty subset with a lower bound has a greatest lower bound, which is why we say "the" real number system, not "a" real number system.
Suppose we only have $\Bbb Q$ and we hope to have an ordered field $F$ with the $glb$ property. Then there cannot exist $b\in F$ such that $0\ne b$ and $|b|\leq 1/n$ for every $n\in \Bbb N.$
Proof: Suppose not. (i). Consider the set $S=\{|b|/n:n\in \Bbb N\}.$ Then $|b|^2$ is a positive lower bound for $S$ because for every $n\in \Bbb N$ we have $|b|^2=|b|\cdot |b|\leq |b|\cdot (1/n).$
(ii). But $S$ has no $glb.$ Because if $c>0$ and $c$ is a lower bound for $S,$ then $c<2c.$ And for every $n$ $\in \Bbb N$ we have $c\leq |b|/2n,$ implying $2c\leq |b|/n.$ So $2c$ is also a lower bound for $S$.
Υou are right . I cannot find any problem in your nice method, but you forgot the absolute value of $b$.
One another approach would be considering $b$ as a sequence...then $lim{1/n}=0$ so by contraction lemma we should also have $limb=0$ which leads of course to $b=0$
