What is the infimum of $\left\{\frac{1}{2^x},x\in\mathbb{N}\right\}$? Prove it. My attempt:
First, $0$ is a lower bound since $0>\frac{1}{2^x}\rightarrow 0>1$, which is impossible.
Assume $\exists r \in (-\infty,0)\in \mathbb{R} $ s.t. $\frac{1}{2^x}\geq r>0, \forall x \in N$
Since $r>0$ then by the archimedean property, $2^x>\frac{1}{r}$ this implies that $r>\frac{1}{2^x}$ so there is an element in the sequence that is greater than $r$ so it is impossible for it to be an infimum or lower bound.
My main concern is with my usage of the archimedean property, I know that the main property states the following:
$$\forall \epsilon>0\ \exists n\in \mathbb{N}:\frac1n<\epsilon$$
however, my logic is that if $n \in \mathbb{N}$ then it will be $n\in \mathbb{R}$ is true as well since $\mathbb{N} \subset \mathbb{R}$, and since it says $\forall \epsilon$ then $\frac{1}{2^x}$ is inside that as well.
 A: Let me dissect your proof in some detail, and touch on the use of the Archimedean property. First, it'd be better to start out with a clear claim of what you're about to prove:

We claim that $0$ is the infimum of the set $S=\{\frac{1}{2^n}:n\in\mathbb N\}$, meaning that it is both a lower bound and that every other lower bound $r$ satisfies $r\leq 0$.

Then you can continue to show that it's a lower bound:

Note that $0 \leq \frac{1}{2^n}$ for every $n$, so $0$ is a lower bound for the set.

I don't think you need to justify it further than that, to be honest. This is the sort of proof you'd do after you've established the basic properties of an order well enough to claim this without justification. However, if you are going to justify it, your proof that "$0 > \frac{1}{2^n}$ implies $0 > 1$ which is a absurd" is an unnecessary use of contradiction - surely you could more easily observe that $0\leq 1$ and that this implies $0 \leq \frac{1}{2^n}$ by diving by $2^n$ (which is positive), creating a direct proof.
You should format the second part of your proof explicitly by contradiction:

Suppose, for contradiction, that there is some $r\in\mathbb R$ which is a lower bound for $S$ and for which $r>0$.

This is probably what you meant to write symbolically, although what you actually wrote was "Assume $r$ is a negative number, which is a lower bound for $S$ and also positive" - which is clearly not what you meant and which will take the reader longer to figure out than if you wrote it in text.
Your idea to apply the Archimedean property is fine, but it's better to write it out more literally, not trying to extract anything about powers of $2$ yet:

Since $r>0$, we can apply the Archimedean property to find some $n\in\mathbb N$ such that $\frac{1}n < r$.

To get a power of $2$ with a similar property, we can then use the $n$ we just received to make a smart choice:

Let $m$ be any natural number such that $2^m \geq n$. Note that $\frac{1}{2^m} \leq \frac{1}n$ hence $\frac{1}{2^m} < r$.

Here we assume that the reader will agree with us that there is some such $m$ (and it's certainly possible to prove this if needed - indeed, you could simply pick $m=n$ if you felt lazy). This step is missing in your reasoning - and handwaving about how $\mathbb N$ is a subset of $\mathbb R$ is irrelevant. The critical properties are that (1) if we're interested in finding $\frac{1}n < r$, it doesn't hurt to make $n$ bigger, and (2) there is always a power of $2$ bigger than any fixed natural number.
At this point we can just conclude by noting:

Note that $\frac{1}{2^m}\in S$, hence $r$ is not a lower bound for $S$, contradicting our assumption. Therefore, to the contrary, every lower bound of $S$ is less than or equal to $0$. Thus $0$ is the infimum of $S$.

