Prove that $B_{0}=\{x\in (0, 1) \mid x=m/2^{n} \;\text{for some} \; m\in\mathbb{N} \;\text{and}\; n\in\mathbb{N}\}$ is countable. Haven't got any clue yet, should we prove there is a bijection between $B_0$ and $\mathbb{N}$ ?
Another attempt:

Let $B:=${$m/2^n| m,n \in \mathbb{N}$}.

$f: \mathbb{N}×\mathbb{N}$ $\rightarrow$ $B$,

$f:(m,n) \mapsto m/2^n.$

$f$ is a surjective map from a countable 

set $ \mathbb{N}×\mathbb{N}$ to $B$.

Hence $B$ is countable.

$B_0 \subset B$, i.e 

$B_0$ as a subset of a countable set $B$ is countable.
Consider the map $f: \mathbb{N}\times \mathbb{N} \to B_0$ given by: $f(m,n) = \dfrac{m}{2^n}$, and $f$ is a bijection and the latter set is countable since former also is.