I want to prove that there exists a rational number between any two real numbers or that the rationals are everywhere dense in the set of reals. I'm taking the set of reals to be an Archimedean ordered field, and by the Archimedean axiom$^{1}$ it is true that for some real number $x$, and positive integers $n,~k$ that $k{1 \over n} > x$, and if there exists a real number $y$ such that $y > x$, then we could say that $y = k{1 + n \over n}$, and therefore $x < k{1 \over n} < y$, and therefore there exists a rational between (positive) reals. The case for $0$ is trivial as there exists a real $x$ and $y$ such that $x < 0 < y$, and for the negative reals we say that for $0 \geq y > x$, then $0 \leq -y < -x$ and so our previous proof for the positives apply, and therefore there exists a negative rational too.
Edit: I realize that I made a mistake by saying $y$ is only equal to $k{{1+n} \over n}$. It should be the case that $y \geq k{1 \over n}$ otherwise $y$ wouldn't be greater than $x$ as $x < k{1 \over n}$. We could still set $y$ such that $y \geq k{1 \over n}$ by necessity, and continue with the proof normally. (Check the comments here: https://math.stackexchange.com/a/4237182/956717)
1: It states that for every real number there exists positive integer which is greater than the real. I.e. $n > x$, which implies that $ny > x$, for some positive $y$ if we replace $x$ by $x \over y$, and if we also replace $x$ in $n > x$ with $1/b$ for some positive $b$ then $x > {1 \over n}$.
I looked at the proof in my book and it seemed very lengthy, and so I thought of a way to shorten it but I'm doubting myself, and I need external validation on whether it is correct or not.
I hope the question wasn't unproductive or the proof isn't embarrassingly wrong. Thanks for helping! (If the answer is correct, please do comment so!)
Since it is to be proven that set of rationals $\mathbb Q$ is dense in $\mathbb R$, you need to prove that given any real numbers $x$ and $y$ such that $x\lt y$, there must be a rational $q\in \mathbb Q$ such that $x\lt q\lt y$.
In your proof, you have arbitrary $x$ but then you chose $y$ such that it is greater than $\frac kn$. This restricts the arbitrary nature of $y$ therefore the proof needs work there.
Alternatively, you could proceed like this:
Let $x\lt y$ be any given real numbers. Hence, $y-x\gt 0$. It follows by Archimedean property that there is an $n\in \mathbb N$ such that $n(y-x)\gt 1\implies ny\gt nx+1$
It follows by the property of floor functions that $\lfloor nx\rfloor\le nx\lt \lfloor nx\rfloor+1\le nx+1\lt ny$, where $\lfloor \rfloor$ is floor function.
It follows that $nx\lt \lfloor nx \rfloor+1\lt ny\implies x\lt \frac{\lfloor nx \rfloor+1}{n} \lt y$.