Is it possible to construct a strictly monotonic sequence of all rational numbers?

if $a_k< a_{k+1}$ then $x := \frac{a_{k+1}+a_k}{2}$ is a rational number between those two, so no.

You are not mistaken. Even assuming you want just the non-negative numbers, so that $a_0=0$, you cannot pick $a_1$ correctly because you will have skipped $a_1/2$.

Alternatively, you could allow negative indices (and have $\ldots, a_{-1}, a_0, a_1,\ldots$), which would also solve your "there is no smallest rational number" problem, but it still has the problem that there are rational numbers between any two numbers. Specifically, if the list is complete, we must have some $n$ such that $a_n=0$. But then we necessarily miss $a_{n+1}/2$.

As Arthur already stated, such a numbering cannot start with a finite index, i.e. either your sequence will count from $-\infty$ to $\infty$ or you only count the non-negative rational numbers (i.e. $\mathbb Q^+_0$). And as Thomas showed, a "normal" sequence also won't work since you'll always find a number in between.

However, you can define a sequence of sequences $\{\{a_{nm}\}_{n=-\infty}^\infty\}_{m=1}^\infty$ such that its $\lim_{m\to\infty}$ yields a sequence counting all rational numbers. As an example, consider the typical enumeration sequence of $\mathbb Q$ (see e.g. here) and let $\{a_{nm}\}_n$ be the ordered sequence of the first m rational numbers obtained that way. The thing is, though, you will just end up with $\mathbb R$...