Questioning the discrete nature of $\mathbb{Q}$ Background for question:
I recently heard someone say : " The measure of the set of rational numbers $\mathbb{Q}$ is $0$. ".
And I thought, well that makes sense because $\mathbb{Q}$ is made up of points. So its measure should naturally be sum of $0$ lengths. 
But then I asked myself : Why did you assume $\mathbb{Q}$ is made up of discrete points? 
I thought there must be an irrational number beside every rational. But then why is that? Why is it 'impossible' to have two rational numbers right next to each other on the number line? 

So my Question:

Conjecture : There must be at least one set $A=\{a,b    \}$ such that $a,b \in \mathbb{Q}$ and there is no number between $a$ and $b$. Also $a \neq b$.
How do we prove or disprove this (In a mathematical way, not on pure intuition) ?


My try and thoughts: 
I couldn't get any valid starting point for this. All I could mange to do was write the conjecture in different format. 
eg. $$\lim_{h \to 0^{+}} a+h =b$$
But I have sort of a hunch that this would be disproved using proof by contradiction. 

All of $\Bbb R$ is made up of points, but there are uncountably many of them.  For $\Bbb Q$ there are only countably many, which is what we use to show the measure is $0$.  We know that Lebesgue measure is countably additive, so if we can cover $\Bbb Q$ by a set of intervals the measure of $\Bbb Q$ must be less.  
Put the rationals in order as $q_1, q_2, q_3, \ldots$.  Then we can make an interval around $q_1$ of length $\frac h{2^1}$, an interval around $q_2$ of length $\frac h{2^2}\ldots,$ an interval around $q_i$ of length $\frac h{2^i}$ and so on.  The total of all these intervals is no more than $h$.  We can make the total as small as we want by shrinking $h$, so the measure of the rationals must be $0$.  
Both the reals and rationals are dense.  Between any pair of them lies another.  There is no concept of reals or rationals being next to each other.
