Help with proof that every point $x \in [0,1]$ is an accumulation point of given sequence I was aksed to show that every point $x \in [0,1]$ is an accumulation point of the sequence
$$
v_n= \sum_{k=0}^Kz_k10^{-k-1} \text{, where } n= \sum_{k=0}^Kz_k10^k.
$$
So for example $v_{123}=0.321$ or $v_{3210}=0.0123$. I think I understand the question and got the idea but I don't quite know how to write down the proof.
Here is my attempt - hopefully not a mess:

Proof. Let $x \in (0,1)$ be arbitrary.The point $x$ can be written as
$$
x=0.x_0x_1x_2x_3 \cdots \text {, where } 0 \leq x_i \leq 9, \forall i \in \mathbb N.
$$
Now define the subsequence
$$
y_k =0.x_0x_1x_2x_3 \cdots x_k.
$$
Then we get $|y_k-x| \lt \varepsilon, \forall \varepsilon \gt 0$ if we choose $N= \lceil \frac{1}{n} \rceil$ and $lim_{k \to \infty} y_k=x$. If now $x_i=z_i, \forall i \in \mathbb N$, we get an $n_k$ with
$$
n_k= \sum_{i=0}^Kz_i10=\cdots x_3x_2x_1x_0 \implies v_{n_k}= \sum_{i=0}^Kz_i10^{-i-1}=0.x_0x_1x_2x_3 \cdots.
$$
Then
$$
lim_{k \to \infty}v_{n_k}=x
$$
and the point $x$ is an accumulation point of the sequence $v_n$.
For $x=0$, define
$$
v_{n_k} =0.000 \cdots 1.
$$
Then $|v_{n_k}-x|=|v_{n_k}-0| \leq 10^{-k+1} \lt \varepsilon$ with $N= \lceil \frac{1}{n} \rceil$.
For $x=1$ define
$$
v_{n_k}=0.999 \cdots 9
$$
Then $|v_{n_k}-x|=|v_{n_k}-1| \leq 10^{-k+1} \lt \varepsilon$ with $N= \lceil \frac{1}{n} \rceil$.
That means every $x \in [0,1]$ is an accumulation point of the sequence $v_n$.
$$\tag*{$\blacksquare$}$$

I'm really struggling with this one for a long time now, so thank you very much in advance!
 A: 
Then we get $|y_k-x|<\epsilon$, $\forall\epsilon>0$ if we choose $N=\lceil\frac{1}{n}\rceil$.

That means $y_k=x$, which is false. You're also not using $N$, and $n$ itself is not defined, so I would call this unclear. I know you meant something like, "given $\epsilon>0$, there is $N\in\Bbb N$ such that for all $k>N$, $|y_k-x|<\epsilon$..." but you need to be more precise (and define $n$ !).
However, $\lim_{k\to\infty}y_k=x$ is true (that's what it means to write $x=0.x_0x_1x_2\cdots$).
The same 'mess' with $N=\lceil\frac{1}{n}\rceil$ appears later, for the $x=0,1$ cases, where $n$ is not defined. And indeed, neither $k$ nor $\epsilon$ are defined either. I suggest you work on tightening your usage of quantifiers, in the right order and in the right way.
Your proof idea is right, but - as you suspected - it's the way it's presented that needs work. You've also made an omission - what if $x=0.560000000\cdots$? Then your $v_{n_k}$ are actually all the same number, for $k>3$, so your proof doesn't show $x$ to be an accumulation point (which by definition requires $x$ to be a limit of distinct elements of the set). However, you can just adapt your proof with the same methods you used for $x=0,1$.
Here's how I might do it (for the cases $x\neq0,1$):

Fix an arbitrary $0<x<1$ and express $x$ in decimal form as: $x=0.x_1x_2\cdots x_nx_{n+1}\cdots$ with $x_i\in\{0,1,\cdots,9\}$ for every $i$. Define, for every $k\in\Bbb N$, $y_k(x)=0.x_1x_2\cdots x_k$. By definition of the $(x_i)_{i=1}^\infty$, $\lim_{k\to\infty}y_k(x)=x$.
Define the sequence of integers $(z_i)_{i=1}^\infty$ via $z_i:=x_i$ for every $i$. Define $k_1$ to be the first natural at which $x_{k_1}\neq0$ (such a natural exists, as $x\neq0$). For $j\in\{2,3,\cdots\}$, assume the natural $k_{j-1}$ has been defined. Then, let $k_j$ be the first natural greater than $k_{j-1}$ for which $z_{k_j}\neq0$ if such a natural exists, and proceed with the induction. If no such natural exists, terminate the induction and redefine $z_{k_{j-1}}$ to be $x_{k_{j-1}}-1$ (note that $x_{k_{j-1}}>1$) and redefine $z_i$ to be $9$, for all $i>k_{j-1}$. For $j'\ge j$, define $k_{j’}:=k_{j-1}+(j'-j+1)$.
In either case, define: $$n_j:=\sum_{i=0}^{k_j}z_i\cdot 10^i$$
We have: $n_1<n_2<\cdots<n_j<n_{j+1}<\cdots$ and $v_{n_1}<v_{n_2}<\cdots<v_{n_j}<v_{n_{j+1}}<\cdots$ for all $j\in\Bbb N$. Moreover, in the case that the induction never terminated: $$v_{n_j}=y_{k_j}(x)$$And as the sequence $(k_j)_{j=1}^\infty$ is an increasing subsequence (tending to infinity) from $\lim_{k\to\infty}y_k(x)=x$ we get: $$\lim_{j\to\infty}v_{n_j}=\lim_{j\to\infty}y_{k_j}(x)=x$$
By definition, as the $(v_{n_j})_{j=1}^\infty$ are distinct, $x$ is an accumulation point of the sequence.
In the case that the induction terminates, say at step $J$, we know that: $$x=0.x_1\cdots x_{k_{J-1}-1}x_{k_{J-1}}$$But we also know that: $$v_{n_j}=0.x_1\cdots x_{k_{J-1}-1}(x_{k_{J-1}}-1)\underset{j-J\text{ times}}{\underbrace{99\cdots99}}$$For all $j\ge J$. Then $|v_{n_j}-x|=10^{-(1+j-J)}$, for all $j\ge J$. Since: $$\lim_{j\to\infty}10^{-(1+j-J)}=0$$It follows that: $$\lim_{j\to\infty}|v_{n_j}-x|=0$$
Again, as the $(v_{n_j})_{j=1}^\infty$ were all distinct (!) $x$ is an accumulation point of the sequence.

The cases $x=0,1$ are quite easy, just consider $0.1,0.01,\cdots$... or $0.9,0.99,\cdots$, but be careful with your quantifiers. Notice that I also used limit notation in preference to $\epsilon$-$N$ notation - it is often cleaner that way, while still carrying the same information.
