Proving an infimum using the archimedean property of R As the title states, I am working on proving an infimum with the archimedean property of $\mathbb{R}$. The problem is stated as follows:

Use the archimedean property of $\mathbb{R}$ to rigorously prove that $inf \{\frac{1}{n} : n \in \mathbb{N}\} = 0$

A note: I realize the below is probably the wrong way to go about it, but it is an attempt - im not great at proofs yet...
I start out by stating that $A = \{\frac{1}{n} : n \in \mathbb{N}\}$
I want to prove that $infA = 0$. 
I feel like a proof by contradiction is a good way to go about it:
So, suppose not, suppose $infA \neq 0$
Then let some $s$ be a lower bound for $A$. 
Following the definition of an infimum, given any lower bound $b$, it follows $s \ge b$.
Since by our premise $infA \neq 0$, then $s \gt 0$.
Now, following the archimedean property, since $s \in \mathbb{R}$ and $s \gt 0$, we can find some $n \in \mathbb{N}$ such that $\frac{1}{n} < s$. 
It follows then that $1 < sn$. But the supremum of the set $A$ is 1. Thus we have reached a contradiction, and it must be that $inf A = 0$.
To be honest I feel intimidated by trying to prove this stuff, and could really use the support. I have ordered a book on proofs, but I feel like im struggling to really "grasp" what is going on beyond just putting stuff together. Aside from helping with the proof itself, I'd appreciate any advice on how to decide when to use proof by contradiction in the future. Thanks!
 A: To prove $\inf\{1/n:n\in\mathbb{N}\}=0$, it must be shown that (i) $0$ is a lower bound and (ii) $0$ is the greatest lower bound.
Proof of (i): Fix $n\in\mathbb{N}$. Observe $0<1/n$. We are done.
Proof of (ii): Let $x$ be an arbitrary lower bound. It remains to show $x\le 0$. By contradiction, suppose $0<x$. Then by Archimedes, there exists $N$ such that $1<Nx$. Therefore $1/N<x$. Therefore $x$ is not a lower bound---a contradiction. As a result, $0\nless x$. Because $\mathbb{R}$ is totally ordered, it follows that $x\le 0$ as desired. We are done.
A: In the context of this problem, the lemma says:
Let $0$ be a lower bound for $A$. Then $0 = $inf $A$ if and only if for every $ε > 0$, there exists $a \in A$ such that $a < 0 + ε$.
To use this Lemma to prove $0 =$ inf $A$, you must show that $0$ satisfies two statements:
$(i)$ $0$ is a lower bound for $A$.
$(ii)$ For every $ε > 0$, there exists $a \in A$ such that $a < ε$.
Explain that this is what you need to show and why. Then prove $(i)$ and $(ii)$ as two separate proofs.
A: You get the logic, but you fall short in formalizing it rigorously.
$A$ is non-empty and bounded below by $0$, since $1/n>0\forall n\in\Bbb N$. So $\inf A$ exists, let it be $s$. 
Since $0$ is a lower bound, $s\ge0$, by the basic definition of infimum. But if $s>0,\exists n\in\Bbb N|1/n<s$ by the Archimedian property. Thus $s\not>0$ because you can always find elements of $A$ smaller than $s$ in that case. Thus, $s=0$.
There is no need to allude to the supremum of $A$.
