I would like to show that there exists a topology $\tau$ on $\mathbb{N}$ such that $(\mathbb{N},\tau)$ is homeomorphic to $(\mathbb{Q}\cap[0,1], \tau_{usual})$.
Sierpinski’s theorem: Any countable zero-dimentional without isolated points is homeomorphic to $\mathbb{Q}$, the rationals with the order topology (same as $\mathbb{Q}$ as a subspace of $\mathbb{R}$ with usual topology, or as $\mathbb{Q}$ with the metric topology).
By Sierpinski's theorem is enough to show ${\cal B}\subseteq [\mathbb{N}]^{\omega}$ closed under complements, and every non-empty intersection is in ${\cal B}.$ Then $(\omega, \tau)$, where $\tau:=\left<\mathcal{B}\right>$ is the topology generated by ${\cal B}$, is homeomorphic to $\mathbb{Q}\cap[0,1].$
I've been worked with the following idea to construct ${\cal B}$: Let ${\cal F}$ be a splitting family for $[\omega]^{\omega}$, i.e. a family ${\cal F} \subseteq [\omega]^\omega$ such that for every infinite $X$ there exists $A \in {\cal F}$ such that $$|X \cap A|=|X \setminus A|=\aleph_0.$$ It's clear that I can suppose that ${\cal F}$ is closed under complements.
My question is: Could I assume that ${\cal F}$ is closed under non-empty intersections?
Just take any bijection $T$ from $\mathbb N$ to $\mathbb Q\cap [0,1]$ and declare $A$ to be open in $\mathbb N$ if $T(A)$ is open in $(\mathbb Q\cap [0,1], \tau_{usual})$. $T$ becomes a homeomorphism.