Consider the set $\mathbb Z\times[0,1]$ and enumerate the rationals $\mathbb Q=\{q_n:n\in \mathbb Z\}$. For every $n$ identify the point $(n,0)$ with $(q_n,0)$. Is there a natural topology (separable metric) on this set so that at height $1$ there is just a discrete point set, and at the height 0 there is the rationals, and each 'vertical' interval is in the standard topology as a subspace?
This is hard to draw so also is there a better way of describing this space?
On
Let $X = \mathbb Z \times [0,1] \cup \mathbb Q \subset \mathbb R^2$ as a set, and define a metric on $X$ by $$ d((x,y), (x',y')) = \begin{cases} |y-y'| &: x=x' \\ |x-x'| + y + y' &:\text{otherwise} \end{cases} \,. $$ I believe this gives your space. It should be easy to check that $X \cap \mathbb Q^2$ is a countable dense subset.
The space is homeomorphic to
Q×{0} $\cup$ Z×{0}.
a subspace of the real plane.