In every interval there is a rational and an irrational number. When the interval is between two rational numbers it is easy. But things get complicated when the interval is between two irrational numbers. I couldn't prove that.

Supposing you mean an interval $(x,y)$ of length $y-x=l>0$ (it doesn't matter whether $l$ is rational or irrational), you can simply choose any integer $n>\frac1l$, and then the interval will contain a rational number of the form $\frac an$ with $a\in\mathbf Z$. Indeed if $a'$ is the largest integer such that $\frac{a'}n\leq x$ (which is well defined) then $a=a'+1$ will do. By choosing $n>\frac2l$ you even get two rationals of this form, and an irrational number between those two by an argument you claim to already have.
