I found the following formulation of the completeness axiom:
If two non-empty sets ($A$ and $B$) are such that for any two elements from the first and second sets ($a$ and $b$, respectively) it is true that $a \le b$, then there is a number $c$ such that $a \le c$ and $c \le b$.
It is also said that this axiom does not hold for rational numbers. However, if we take two non-empty sets ($A$ and $B$) such that $a \le b$ and find among the elements of the two sets such $a$ and such $b$ that $a = b$, then we can say that we found such $c$ ($c = a = b$) that $a \le c$ and $c \le b$. So, for that case it seems to me that the axiom holds also for rational numbers.
Shouldn't it be the case that in the formulation of the axiom we should requests that for any $a$ and $b$ it should be the case that $a < b$ (instead of $a \le b$)? But even in this case we can also find another rational number $c$ (c = (max(a) + min(b)) / 2) such that $a \le c$ and $c \le b$.
So, it looks to me that the completeness axiom holds for rational numbers if we stay in the domain of rational numbers. Do I misunderstand something?
While between any two numbers there is a rational number, this doesn't save completeness. Let $A=\{q\in\mathbb{Q}: q^2<2\}$ and $B=\{q\in\mathbb{Q}: q^2>2,q>0\}$. Then every element of $A$ is less than every element of $B$, but there is no rational which is simultaneously $\ge$ every element of $A$ and $\le$ every element of $B$.
The point is that completeness looks at the interaction between sets of numbers, not just individual numbers themselves. So while $\mathbb{Q}$ and $\mathbb{R}$ look similar in some ways (namely, each is a dense linear order without endpoints), completeness totally breaks for $\mathbb{Q}$ but holds for $\mathbb{R}$.
This "points-to-sets" shift can be subtle at first, and easy to miss. In your OP for example you start talking about sets $A$ and $B$ but wind up talking about individual numbers $a$ and $b$, and this is what makes it look like some semblance of completeness holds in $\mathbb{Q}$: while for any specific $a\in A$ and $b\in B$ there will be a rational number between them, there is no rational number which is between the sets $A$ and $B$ in the appropriate sense - no single rational $c$ which works for all possible choices of $a$ and $b$.