Assume $y-x>0$. By the Archimedean property, there is $n$ such that $n(y-x)=ny-nx>1$. But if two reals have a gap $>1$; there must be an integer between them. So...?
For irrationals, pick your favourite one in $(0,1)$ and transport it between two rationals by an appropriate map. That is, if $\kappa$ is irrational, and $r<s$ rationals, can you find rationals $\alpha\neq 0 ,\beta$ such that $r<\alpha\kappa+\beta<s$? Then use the first part to conclude. Note $\alpha\kappa+\beta$ is still irrational! (Why?)
The proof that $\mathbb{R}$ does indeed have the Least Upper Bound property really depends upon how you're defining the real numbers; for example, if $\mathbb{R}$ is constructed using Dedekind cuts, then the proof is rather straight-forward and easy. If you're constructing $\mathbb{R}$ using equivalence classes of Cauchy sequences, then it's involved (at least the proofs I've seen/done).
Thus, I'll use this answer to address the concerns of why the Least Upper Bound property needs to be mentioned at all, and why it doesn't trivially hold true for everything.
Firstly, the Least Upper Bound property is essentially the reason calculus can be done; as we shall see, there are "gaps" in the rational numbers. The ability to take limits, which is central to everything done in Real Analysis, is closely related to the Least Upper Bound property. To show why this is the case, I'll quote some equivalences:
- Least Upper Bound Property
- Bolzano-Weierstrass Theorem (all cauchy sequences are convergent) and the Archimedean property
- Monotone Convergence Theorem and the Archimedean property
- Nested Intervals Theorem
All of the above are equivalent, and all are central to Real Analysis.
As far as the Least Upper Bound property not holding more generally in, for example, $\mathbb{Q}$, consider the following set: $$\{ q\in \mathbb{Q} \mid q>0 \wedge q^2<2 \}$$ It is clearly bounded above; 2 is an upper bound, for example. Does this set have a least upper bound? In $\mathbb{R}$ it certainly would, and would be $\sqrt{2}$; since $\mathbb{Q}$ is dense in $\mathbb{R}$, if there were a least upper bound in $\mathbb{Q}$ for this set, then it would also be a least upper bound of the set in $\mathbb{R}$. But we know that $\sqrt{2}$ is irrational, so this cannot be the case. Thus, $\mathbb{Q}$ cannot have the Least Upper Bound property. This is why the Real numbers are necessary.
Best Answer
Let $a<b$ be any real numbers. By Archimedean property there exists a natural number $n$ such that $n(b-a)>1$. It follows that $2^n(b-a)>1$, so the interval $(a,b)$ has length bigger that $\tfrac {1}{2^n}$ and so it contains a dyadic rational points of the form $\tfrac {k}{2^n}$ for some integer $k$, because such points are placed at the real line with a distance $\tfrac {1}{2^n}<|b-a|$ between consecutive points. Similarly we can show that the interval $(a,b)$ contains an rational point of the form $\tfrac {k}{2^n}\cdot \tfrac{1}{\sqrt 2}$ (with a little care assuring that $k\ne 0$).