I know about the proof found here: Proof there is a rational between any two reals.
I wanted to know if this similar proof is also correct?
Assume $x > 0$. Since $y > x$, it follows $y-x>0$. There exists some $n\in \mathbb{Z}^+$ such that $\frac{1}{n}<\min(y-x,x)$. Define $B=\{\frac{k}{n}\mid nx\ge k\in\mathbb{Z}^+\}$ which is nonempty by construction and bounded above by $x$.
Thus, $\sup(B)=\beta$ exists and $x-\beta<\frac{1}{n}$ by construction. Furthermore, $\beta\in B$ by the well ordering principle (edit: this is wrong). Hence, there exists some $k\in\mathbb{Z}^+$ so $\frac{k}{n}=\beta$ where $\frac{k+1}{n}>x$. Observe $x<\frac{k}{n}+\frac{1}{n}<x+(y-x)=y$.
If $y<0$, use $x=-y$ and $y=-x$ in the above proof to get $\frac{-(k+1)}{n}$ as the rational between x and y. If $y>0$ and $x<0$, use $0$ as the rational.
Best Answer
Proof looks fine although it's a bit circuitous to get the correct bounding number. You also want to start with "assume $0<x<y$" outright.
Note that $B$ is simply a finite bounded set so it easily contains its supremum. Also probably easier to write "$\exists k\in\mathbb{Z}^+$ such that $$\frac{k}{n}\leq x<\frac{k+1}{n}\text{".}$$
As mentioned in the comment, it's easier to look at $y-x$ from the beginning, and just scale it up large enough: consider $$m=\left\lceil\frac{1}{y-x}\right\rceil+1\text{,}$$ and notice that some fraction $\frac{n}{m}$ must lie strictly between $x$ and $y$, since $x+\frac{1}{m}<y$, therefore $n=\left\lceil mx\right\rceil+1$ works.