Joke (but true): The difference between a rational number and an irrational number is irrational.
Serious answer: Your question already expressed it. A rational number can be written $\frac mn$ for some integer $m$ and some positive integer $n$. An irrational number is a real number that cannot be written like that.
To show that a number is rational, the most common approach by far is to find $m$ and $n$, and prove that the number in fact equals their ratio.
To show that a number is irrational is often a good deal harder, and is usually done using some sort of proof by contradiction. For example, it took a long time for mathematicians to even prove that $\pi$ is irrational. According to https://mathoverflow.net/questions/40145/irrationality-of-pie-pipi-and-epi2, no one even knows whether $\pi^{\pi^{\pi^\pi}}$ is an integer, let alone whether it is rational (but just about anyone would bet that it's irrational).
It turns out that in several senses, almost all real numbers are irrational, and in fact even transcendental (a nastier sort of beast). There are also various techniques available for manufacturing great gobs of irrational (and even transcendental) numbers, but most of the numbers people are actually interested in are either trivially rational, trivially algebraic (not transcendental), or mysterious—no one knows for certain whether they are rational or irrational.
Part of the reason for this is that while it's very easy to put together rational numbers to get more rational numbers, you can't really put together irrational numbers to get more irrational numbers in very many ways. For example, the sum or product of two rational numbers is always rational, but the sum or product of two irrational numbers may be rational.
Your proof looks fine, but there is a slightly more elementary way of doing this: $\sqrt{2}$ is irrational. A rational times an irrational is irrational, and the sum of a rational with an irrational is irrational. It's then easy to check that
$$s = a+\frac{\sqrt{2}(b-a)}{2}$$
is an irrational number between $a$ and $b$.
Best Answer
Let $x,y\in\mathbb{R}$, $x\neq y$. Without loss of generality, suppose $x<y$. Then there exists a positive $z$ such that $y-x=z$.
By Archimedes' axiom, there exists a natural number $n$ such that $$n > \dfrac{1}{z}$$ $$nz > 1$$ $$ny - nx > 1$$ So there exists an integer $m$ such that $$nx < m < ny$$ $$x < \frac{m}{n} < y$$ i.e. $m/n$ is a rational number between $x$ and $y$.
Since $x$ and $y$ can be any real numbers, in particular they can be irrationals.