Infinite Dyadic Rationals in any open interval $(a,b)$ where $a
Given that there exists at least one dyadic rational of the form $2^{-n}m$ between any two distinct real numbers $a<b$, show that there infinite such rationals between $a$ and $b$.
My Attempt
Given $a<b\in\mathbb{R}$ we have $$a<\frac{m_1}{2^{n_1}}<b,$$where $n$ is a natural number. But then $a<(b+a)/2<b\in\mathbb{R}$ so that $$a<\frac{m_{21}}{2^{n_{21}}}<\frac{b+a}{2}\text{ and }\frac{b+a}{2}<\frac{m_{22}}{2^{n_{22}}}<b.$$So there are at least two dyadic rationals between the reals $a$ and $b$. However, $$a<\frac{a+(b+a)/2}{2}<\frac{b+a}{2}\text{ and } \frac{b+a}{2}<\frac{(b+a)/2+b}{2}<b.$$By applying the given result again, we have $$a<\frac{m_{31}}{2^{n_{31}}}<\frac{a+(b+a)/2}{2}\text{ , }\frac{a+(b+a)/2}{2}<\frac{m_{32}}{2^{n_{32}}}<\frac{b+a}{2}$$and $$\frac{b+a}{2}<\frac{m_{33}}{2^{n_{33}}}<\frac{(b+a)+b/2}{2}\text{ , }\frac{(b+a)/2+b}{2}<\frac{m_{34}}{2^{n_{34}}}<b.$$ So, there at least four dyadic rationals in $(a,b)$. 
After the $k$-th iteration we will have partitioned the interval $(a,b)$ exactly $2^{k-1}$ times. Thus, we will have reached the conclusion that there exist at least $2^{k-1}$ dyadic rationals in $(a,b)$. Repeating the process indefinitely shows that there exist $$\lim_{k\to\infty}2^{k-1}=\infty$$dyadic rationals in $(a,b)$. 
Note: the first number in the indices stands for the number of intervals.

A PhD student in a support class we have for Analysis told me that my proof is a bit flawed because as $k$ grows large then the intervals in which I have partitioned $(a,b)$ become smaller and smaller and tend to zero.  If at infinity they become zero, then there cannot be a rational dyadic (or any number for that matter) inside those intervals, and hence my proof is invalid. He didn't seem to be sure about this though but he suggested that I try a different proof altogether so that I avoid this issue. 
In my opinion, his assertion is not so sound, because this would imply that because intervals get smaller and smaller then all the intervals are countable, which is not the case. However, I haven't yet been taught anything on countability of sets, so I hope you can enlighten me on the matter. 
Thanks in advance!
 A: Your proof is fine except for some informal "and so on" in it. 
Even if the interval lengths tend to zero, they never are zero. Instead of invoking the $\lim$, however, it may be better to state that $2^{k-1}$ exceeds any natural number $N$ (then again, this is exactly what $\lim 2^{k-1}=\infty$ says), hence the assumption that there are only $N$ such numbers leads to a contradiction, no matter how big $N$ is.
A variation of the theme might be this:
Let $S$ be the sets of dyadic numbers in $(a,b)$.
Assume $|S|$ is finite.
As there is at least one dyadic in $(a,\frac{a+b}2)$ and one in $(\frac{a+b}2,b)$, we conclude $|S|\ge 2$ so that we can 
let $x=\min S$ and $y=\min(S\setminus\{x\})$. [This uses that every finite nonempty set of reals has a minimal element.] Then $S\cap (x,y)=\emptyset$, but as $x<y$ there exists a dyadic $z\in(x,y)\subseteq (a,b)$, hence $z\in S$ - contradiction.
A: I think your proof is conceptually fine, with the main "flawish" part formalized by Hagen von Eitzen. Perhaps, more easily, you can also start with the assumption that there are at most $N$ such numbers, divide $(a,b)$ into $N+1$ intervals of equal length (this way they obviously become intervals of non-zero length) and obtain a contradiction.
