Q:

My Proof for the Cardinality of a Particular Binary Distribution

my question reads as follows:

I have constructed a proof and am concerned about 2 things:
1) The  validity of my proof.
2) The construction of my proof.
I am asking for someone to read through it and tell me where I can improve my proof for presentation.

Proof:
The set $W$
  is uncountable. 
Assume $W$
  is countable. By definition, $W$
  is countable if there exists a function, $f:W\to\mathbb{N}$
 , such that $f$
  is a bijection. Furthermore, countability implies that the elements of $W$
  can be arranged as a list. Suppose that all the possible elements of $W$
  can be expressed as follows:
$\omega_{0} = x_{00},x_{01},x_{02},x_{03},x_{04},.....$
$\omega_{1} = x_{10},x_{11},x_{12},x_{13},x_{14},.....$
$\omega_{2} = x_{20},x_{21},x_{22},x_{23},x_{24},.....$      (1)
$\omega_{3} = x_{30},x_{31},x_{32},x_{33},x_{34},.....$
$. . .
. . .$
,where $\omega_{i}\in W$
 , $x_{ij}\in\{0,1\}$
  for $i,j\in\mathbb{N}$
 . In addition, the restriction on $W$
  implies that
$x_{ij}+x_{i(j+1)}+x_{i(j+2)}\neq\begin{cases}
3 & ,\ as\ (x_{ij},x_{i(j+1)},x_{i(j+2)})=(1,1,1)\\
0 & ,\ as\ (x_{ij},x_{i(j+1)},x_{i(j+2)})=(0,0,0)
\end{cases}$,        (2)
for all $x_{ij}\in\{0,1\}$
 . Assuming $W$
  is countable implies that our bijective function, $f$
 , is defined as $f(w_{i})=i$
 .
Let $\tau_{n}=x_{n0},x_{n1},x_{n2},x_{n3},x_{n4},\cdots$
  for $n\in\mathbb{N}$
  , be an infinite binary adhering to condition (2)
 . Let $x_{n0}=\{0,1\}\backslash\{x_{00}\}$
  and $x_{n1}=\{0,1\}\backslash\{x_{11}\}$
 . This implies that $\tau_{n}$
  is distinct from binaries $\omega_{0}$
  and $\omega_{1}$
 . In determining the value of $x_{n2}$
 , caution must be taken to not only create a distinction from $\omega_{2}$
  but to also satisfy condition (2)
 . Thus we let
$x_{n2}=\begin{cases}
\{0,1\}\backslash\{x_{22}\} & ,\ if\ x_{n0}+x_{n1}+\{0,1\}\backslash\{x_{22}\}\neq3\ or\ 0\\
x_{22} & ,\ if\ x_{n0}+x_{n1}+\{0,1\}\backslash\{x_{22}\}=3\ or\ 0
\end{cases}.$
If $x_{n0}+x_{n1}+x_{22}\neq3$
  or $0$
 , then $(2)$
  has been satisfied and $\tau_{n}$
  is distinct from $\omega_{3}$
 . Thereafter, proceed to make a distinction from $\omega_{4}$
  using the same procedure. $If x_{n0}+x_{n1}+x_{22}=3$
  or $0$
 , then while $\tau_{n}$
  has satisfied $(2)$
  it is not distinct from $\omega_{2}$
 . Thus, continue to find a distinction from $\omega_{2}$
  by iterating this process for $x_{23}$
 .
That is,
$x_{n3}=\begin{cases}
\{0,1\}\backslash\{x_{23}\} & ,\ if\ x_{n1}+x_{n2}+\{0,1\}\backslash\{x_{22}\}\neq3\ or\ 0\\
x_{23} & ,\ if\ x_{n1}+x_{n2}+\{0,1\}\backslash\{x_{22}\}=3\ or\ 0
\end{cases}.$
Continue this process until a distinction from $\omega_{2}$
  is made. That is,
$x_{np}=\begin{cases}
\{0,1\}\backslash\{x_{3p}\} & ,\ if\ x_{n(p-2)}+x_{n(p-1)}+\{0,1\}\backslash\{x_{3p}\}\neq3\ or\ 0\\
x_{3p} & ,\ if\ x_{n(p-2)}+x_{n(p-1)}+\{0,1\}\backslash\{x_{3p}\}=3\ or\ 0
\end{cases}$.
Once a distinction has been made proceed to make a distinction from $\omega_{3}$
  following the same process above. Continuing this overall iteration creates a binary $b_{n}$
  such that it is not contained in the list, $(1)$
 . Therefore $W$
  can not be listed and thus is uncountable. $\square$

Thank you all in advanced for you time !

A:

Two suggestions: 
(a) Show that eventually, there will be a distinction.
(b) Once you assume that $W$ is countable, you can go on to say that $W$ can be expressed as $\ldots$ (Also, feel free to use $w$ instead of $\omega$, but this doesn't matter.) On a similar note, giving a bijection $f: W \to \mathbb{N}$ is exactly the same thing as listing the elements of $W$.
However, if I wanted to prove your problem, I'd want to use one of the following: 
Lemma 1
Assume $A$ is uncountable. If there is an injection $A \hookrightarrow B$, then $B$ also is uncountable.
or 
Lemma 2
Assume $B$ is uncountable. If there is a surjection $A \to B$, then $A$ also is uncountable.
For lemma 2, $B$ should could be the collection of all sequences of 0's and 1's, which is uncountable. In your case, $W$ is $A$. For lemma 1, this is swapped, and applying lemma 1 should be easier then lemma 2.

