real analysis - How to pick decimal expansion in the proof that $(0,1)$ uncountable

Prove $(0,1)$ is uncountable.

Suppose $(0,1)$ were countable.

List $(0,1)$ as:

$x_1=0.a_{11}a_{12}\dots$

$x_2=0.a_{21}a_{22}\dots$

and so on,

where $a_{ij}$ are integers from $0$ to $9$.

Let $b_n=3$ if $a_{nn}=4$ and $b_n=4$ if $a_{nn}\neq4$.

Given $0.b_1b_2\dots,$ there must exist some $n$ where

$$x_n=0.b_1b_2 \dots b_n\dots=a_{n1}a_{n2}\dots a_{nn}\dots$$
Then we would say oh but then $b_n=a_{nn}$ and this is a contradiction.

This seems to depend on decimal representations being unique, but I do not know how to pick our expansions $a_{ij}$ such that we can guarantee that the $0.b_1b_2\dots$ that we construct, is the same as one of the $a_{ij}$ in terms of decimal representation.

Answer

Non-unique decimal representations can only occur when the last digits are either $0$ or $9$. For example, $0.4\overline{999}=0.5\overline{000}$.

Choosing $4$ and $3$ for the diagonal number ensures that the representation is unique.