Prove that the dyadic rationals $D$ are dense in $\mathbb{Q}$ 
Prove that the dyadic rationals $D$ are dense in $\mathbb{R}$.

I saw some proofs using showing that $D$ is dense on $[0,1]$.
I proved that if $A$ is dense in $B$ and $B$ is dense in $C$, then $A$ is dense in $C$. So, I want use it. My idea is show that $D$ is dense in $\mathbb{Q}$. Can someone help me? Just a hint, not a solution.

Attempt. Let $D = \left\{\frac{m}{2^{n}}\mid n \in \mathbb{N}\text{ and }m \in \mathbb{Z}\right\}$. Take $I = \left(k-\frac{\epsilon}{2},k+\frac{\epsilon}{2}\right)$ an open interval in $\mathbb{Q}$. We can find a $n \in \mathbb{N}$ such that $\epsilon > \frac{1}{2^{n}}$. If there is no $\frac{m}{2^{n}} \in I$, there is $\frac{m}{2^n}$ and $\frac{m+1}{2^{n}}$ such that $\frac{m}{2^{n}} < k - \frac{\epsilon}{2}$ and $k + \frac{\epsilon}{2} < \frac{m+1}{2^{n}}$. Thus
$$\frac{1}{2^{n}} =\frac{m+1}{2^{n}} - \frac{m}{2^n} > \epsilon,$$
a contradiction.
 A: Instead of reducing the statement to the density of $\mathbb{Q}$, just use the same proof idea. 
Remainder: $\mathbb{Q}$ is dense, because given an interval of length $\varepsilon> 0$, it is possible to pick an integer $n\in \mathbb{N}$ such that $n> \frac{1}{\varepsilon}$. Equivalently, $\varepsilon> \frac{1}{n}$. So given the gap of length $\frac{1}{n}$ (the interval), you cannot step over it if the length of your step is $\frac{1}{n}$. Starting from zero, and walking towards the interval, you will step into it at a point. If this requires $k$ steps, then $\frac{k}{n}$ is in the interval. 
Can you modify this proof to make it work for dyadic rationals?
A: Hint: There is no need to go through $\mathbb Q$. Just prove that if $y>x$ there are integers $n,m$ such that $y-x>m2^{-n}$. 
A: The real number $\mathbb{R}$ and hence all subsets of the real numbers have come with a notion of distance $\left(a, b\right) \to \left|a - b\right|$ .
In particular the rationals themselves form a metric space, see ProofWiki .
This means we can use the definition of dense set from Wikipedia that requires the existence of a metric.
Here's the definition of the closure of a set, slightly rewritten
$$ \bar{A} \stackrel{def}{=}  A \cup \left\{\lim_{n \to \infty} a_n \text{ where } a \in \text{Seq}\left[A\right] \right\} \tag{1} $$
Note that $\bar{A}$ is constrained to be a subset of $\mathbb{Q}$ because $\mathbb{Q}$ is where we're taking limits.
$$ A \text{ is dense in } X \stackrel{def}{\iff} \bar{A} = X \tag{2} $$
One way to show it directly is to come up with an explicit sequence of dyadic rationals that converges to a given rational number $\frac{a}{b}$ for every such $\frac{a}{b} \in \mathbb{Q}$ .
