If A is an uncountable subset of the reals such that all elements of A are irrational, is there a countable subset Z of A such that for all x,y in A, there is a z in Z such that x>z>y?
In general, I want to claim that for any uncountable subset B of the reals, there is always a countable subset Y such that $\forall b,b' \in B$ such that $b>b'$, $\exists y\in Y$ such that $b>y>b'$.
As noted in the comments, this is not always possible, because $A$ can contain points that adjacent in $A$. However, it cannot contain too many of them.
Let $A\subseteq\Bbb R$ be uncountable. Let
$$S=\left\{x\in A:\exists y\in A\big(x<y\text{ and }(x,y)\cap A=\varnothing\big)\right\}\,,$$
the set of elements of $A$ that have immediate successors in $A$. For $s\in S$ let $s^+$ be the immediate successor of $s$ in $S$. The intervals $(s,s^+)$ for $s\in S$ must contain distinct rationals, so $S$ is countable.
$\Bbb R$ is hereditarily separable, so $A$ has a countable subset $D$ that is dense in $A$ in the subspace topology on $A$. Suppose that $x,y\in A$ and $x<y$. If $(x,y)\cap A=\varnothing$, then $x\in S$, and $y=s^+$. Otherwise, $(x,y)\cap A$ is a non-empty open subset of $A$ in the subspace topology on $A$, so there is a $z\in D$ such that $x<z<y$.
This is the best that you can do along these lines unless you begin by requiring that $A$ be densely ordered.