I think I can do this first: index all elements of $A$ by the natural numbers, make a map $f$, first send $a_0$ to any rational number, call it $q_0$. Then inductively, depending on the ordering of $a_n$ with $a_0, \cdots, a_{n-1}$, make $q_n:=f(a_n)$ any rational so that the ordering of $q_1, \cdots, q_n$ preserves that $a_1, \cdots, a_n$, which I can always achieve since $\mathbb{Q}$ is dense. And if I ask myself what is $f(a_n)$ I can always find out in $n$ time. So $f$ should be well defined.
So $f(A)$ is a dense subset of $\mathbb{Q}$ without a maximum or a minimum. And I think now it suffices to show that any countable dense subset of $\mathbb{Q}$ without a maximum or a minimum is isomorphic to the whole of $\mathbb{Q}$. Any thoughts?