Difficulty understanding a proof for the existence of a rational between any two real numbers

I am trying to comprehend a proof for a theorem that states between any $x, y \subset{\Bbb{R}}$, there exists $q\subset{\Bbb{Q}}$ such that $x
The proof in question is as follows:

Choose n such that $\frac{1}{n}< y-x$. Consider multiples of $\frac{1}{n}$ which are unbounded. Choose the first multiple such that $\frac{m}{n}>x.$ We claim that $\frac{m}{n}
If that is not the case, we would have $\frac{m-1}{n}y$. but these imply that $\frac{1}{n}>y-x,$ which is a contradiction.

I understand that the first statement follows from the Archimedean principle, but I cannot grasp the intuition for the rest of the proof.

Most of my confusion arises at the last line, starting from how we arrived to the conclusion that $\frac{m-1}{n}
The latter part of the proof is a proof by contradiction where the claim they've made is that $\frac{m}{n}>x$. Note that it says to choose the first multiple $m$ such that the claim holds; then if thats the case any multiple $k$ smaller than $m$ should result in $\frac{k}{n}y$, we can finish the proof algebraically.

Personally, I think that the concept behind the official proof is sound, but that the proof was written in a confusing manner.

Choose $~n \in \Bbb{Z^+}~$ such that $~n > \dfrac{1}{y-x} \implies$

$y - x > \dfrac{1}{n} \implies y > x + \dfrac{1}{n}.$

Choose $~\displaystyle k \in \Bbb{Z}~$ such that $~k \leq nx < k+1 \implies $

$\displaystyle \frac{k}{n} \leq x < \frac{k+1}{n} = \frac{k}{n} + \frac{1}{n} \leq x + \frac{1}{n} < y.$

Actually, this is exercise 3.12.6 from chapter I (the introductory chapter) in "Calculus" 2nd Ed., volume 1, 1966 (Tom Apostol). Preceding this problem is

Theorem I.29 which proves that for any (finite) real number $~r,~$ there exists $~n \in \Bbb{Z^+},~$ such that $~n > r.$

Exercise 3.12.4 from chapter I that asks the math student to prove that for any real number $~r~$ there exists (unique) $~n \in \Bbb{Z}~$ such that $~n \leq r < n+1.$

To give you an example, think of the integers number line. Positive direction is to the right.

Suppose I claim that the multiple of 3 nearest in the positive direction of 4 is 6. That is, between 4 and 6 , there is no multiple of 3. Then it follows that the multiple of 3 just less than 6 has to be smaller than 4. That is the exact conclusion your proof arrived at to showcase the arising contradiction.

Once $\frac{m}{n}$ has been chosen as the first multiple of $\frac{1}{n}$ that is greater than $x$ (here "first" means "least"), you then use the fact that $\frac{m-1}{n}$ is a smaller multiple of $\frac{1}{n}$ to conclude that $\frac{m-1}{n}$ is not greater than $x$: if it were, then $\frac{m}{n}$ would not have been the least.

In other words, $\frac{m-1}{n} \le x$ (your post is somewhat sloppy about the difference between strict and nonstrict inequality).