Prove that $\mathbb{Q}$ is dense in $\mathbb{R}$

It is true that for any real $x$ and $y$ with $x<y$ there is a rational number between them, but we can't prove that from the fact that for any real $x$ and $y$ with $x<y$ there is a real number between them, because that real number between them is not guaranteed to be rational.

As has been pointed out by now (especially in J.W. Tanner's accepted answer) if you have the fact that between any two real numbers, $x$ and $y$ there will be a real number between them, you can't use that fact to claim there will always be a rational number between them as you have no guarantee the number you proved does exist is rational.

This post is to extend how one would show a rational must exist.

The most basic aspect about the real numbers is that for every real number $x$, whether it is rational or irrational, the will be an infinite sequence of rational numbers $q_i$ that will get as close to $x$ "as you like".

This is subtle but it is basic. I'm not entirely certain how Apostol introduced this. But I think the must basic and comprehensible explanation would go something like this:

$x$ lies somewhere in the real numbers. We can divide the real numbers up into intervals of length $\frac 1n$ for some integer $n$ and $x$ will have but be on some interval between $\frac a{n}$ and $\frac {a+1}n$ because ... well, $x$ has to lie somewhere and everywhere will be in one of these intervals. We can make this intervals more and more precise by taking larger and larger values of $n$. If we look as these rational number endpoints and view them as a list we will have an infinite number of rational numbers "honing in" on our real number $x$.

So with that in mind the proof is simple. For $x$ and $y$ there will be a real number $z$ between them. Now $z$ might not be rational, but there are infinite number of rational numbers "honing in" on $z$. Pick one that is closer to $z$ than $z$ is to either $x$ or $y$. That rational number will be between $x$ and $y$.

If that proof seems way to casual and informal... well, probably everyone will agree with you. But we can formalize it up:

Between $x$ and $y$ so that $x< y$ there is a real $z$ so that $x < z < y$. $z$ is real so somehow Apostol must have proven or defined or declared by axiom that there are several rational $q_n \to z$, or in other words, for any value $\delta > 0$ we can find a rational $q$ so that $|q-z| < \delta$.

That might seem complicated and technical but in every textbook on real numbers, it has to have been stated somewhere in some words, maybe very different looking words, that every real number is arbitrarily close to several rational numbers. That's all this is saying. For any distance $\delta$ we can find a rational number that is within $\delta$ of $z$.

So, $x < z < y$ so $y-z> 0$ and $z-x > 0$ and so if we take $\delta = $ the smaller value of $z-x$ or $y-z$ there will be a rational number, $q$ within that distance to $z$. So $z-(z-x) < q < z + (y-z)$ or in other words $x < q < y$.