How to demonstrate that a sequence has subsequences converging in all points of $[0,1]$? The sequence is:
$S = (\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4}...)$
I want to prove that $\forall \alpha \in [0,1]$, there exists a subsequence of $S$ which converges to $\alpha$.
First, I tried the following argument to say that there is a subsequence converging to every rational number $\alpha \in (0,1)$:
If $\alpha \in \mathbb{Q}$, then $\alpha$ can be written as a proper fraction $\frac{a}{b}$, with $a<b$ Then let $k = b - a$, such that $$\alpha = 1-k\frac{1}{b}$$
Every term of the sequence has the form $$s = 1 - l \frac{1}{n}$$
Where $n \in \mathbb{N}$ and $l = 1,2,3, ... , n-1.$
It follows that for $n = b$  and $k = l$, $s$ is equal to $\alpha$. The subsequence formed by this $s$ and all other fractions which can be simplified to $s$ is constantly equal to  $\alpha$:
$$\frac{a}{b},\frac{2a}{2b},\frac{3a}{3b}, ...$$
Naturally, the subsequence $(\frac{1}{n})$ converges to $0$ and the subsequence $(1-\frac{1}{n})$ converges to 1.
So there are subsequences converging for all rationals in [0,1].
How could I prove that the sequence converges for all irrationals in the interval?
Apart from that, if someone has any suggestions on how I could improve my argument, these would be also welcome.
 A: Mark Joshi's argument de-binarised: Take a real number in its infinite decimal expansion  as
$b=0.14159\dots$.
Using its decimal digits define the sequence as below:
$$b_1=\frac1{10},b_2,=\frac{14}{100}, b_3=\frac{141}{1000},b_4=  \frac{1415}{10000},b_6=\frac{14159}{1000000},\ldots,$$
This is a subsequence of yours and converges to $b$.
A: Hint: Denote your sequence by $(a_n)_{n\geq1}$. Let $\alpha\in[0,1]$.
For every $k\in\mathbb{Z}_{\geq2}$, choose $i\in\{1,\ldots,k-1\}$ such that
$$
\left|\frac{i}{k}-\alpha\right|\leq\frac{1}{k}\tag{1}
$$
(show that such an $i$ exists) and take $n_k\in\mathbb{Z}_{\geq1}$ such that $a_{n_k}=\frac{i}{k}$.
Then $(n_k)_{k\geq2}$ will be strictly increasing and, in view of $(1)$, $(a_{n_k})_{k\geq2}$ will be a subsequence converging to $\alpha$.
Existence of $i$: (Geometrically the existence of such an $i$ should be clear)

 In order to choose $i$ such that $(1)$ will be verified, notice that $(1)$ is equivalent to $|k\alpha-i|\leq1$. Now, since $\left|k\alpha-\lfloor k\alpha\rfloor\right|\leq1$, one would like to choose $i=\lfloor k\alpha\rfloor$. The two situations when this choice will fail are the following. First, it may happen that $\lfloor k\alpha\rfloor<1$, in which case $i=1$ works. Hence in order to take this situation into account, we take $i=\max\{\lfloor k\alpha\rfloor,1\}$. Second, it may happen that $\lfloor k\alpha\rfloor>k-1$. In this latter case, $i=k-1$ works. Hence, all in all, it suffices to take $i:=\min\{\max\{\lfloor k\alpha\rfloor,1\},k-1\}$.

A: it contains all finite binary expansion since it contains all numbers of the form $x/2^k.$ 
Any number in $(0,1)$ has an infinite binary expansion since we can always end in an infinite string of 1s. 
Now just take the subsequence corresponding to the finite truncations of this infinite expansion.
