Number coloring

Some numbers might have different decimal notations - they either end with infinitely many zeros or infinitely many nines. If so, we pick the representation with infinitely many zeros.

Set up the binary relation where $x \sim y$ if the decimal representation of $x$ and $y$ differ by a finite number of digits.
It is easy to verify that this is an equivalence relation since it is reflexive, symmetric, and transitive.
In each equivalence class, by the axiom of choice, we can choose a representative element, which we will denote by $a_k a_{k-1} a_{k-2} \dots a_1 . a_{-1} a_{-2} \dots$ .
For this equivalence class, given any other element $b = \ldots b_2 b_1 . b_{-1} b_{-2} \ldots$ , color it with $( \sum_{-\infty}^{\infty} a_n - b_n ) \pmod{10}$ , where we pad with leading zeros.
By construction, since the infinite sum has a finite number of non-zero terms, hence the sum exists and is an integer.
Then it is easy to check that every two numbers in this subset differing by exactly 1 digit have different colors.
Hence, every two real numbers that differ by exactly 1 digit have different numbers.

Gray's Reflected Binary code is used to sequence binary numbers such that only one bit changes every time you increment. (Originally designed to prevent misleading output from electromechanical switches while incrementing.)

You can apply the same pattern to a decimal sequence, even an infinite one. 001,002,003,004,005,006,007,008,009,019,018,017,016,015,014,013,012,011,010,020,021... 0.01,0.02,...

If only one digit ever changes per step the sequence can be coloured using 10 colours.

Let each real number be classified by a unique "header" and "tail". The header is finite, while the tail is infinite. For example, every number written as m/(2^a * 5^b) (for integers m,a, and b) has a tail of 00000..., while m/(2^a * 5^b * 9) might have a tail of 11111..., 22222..., 33333..., up to 88888.... The tail for an irrational number like pi is harder to describe, but it still exists.

Group every number by its tail, and choose a number (O) from each group and set its color to zero. Then, for each number (N) go through the digits in the header and add the difference between the digit in N and the digit in O to the color. For any other number that differs from it by a single digit, the sum will be different. Then your done!

For example, 1/16 (0.0625 00000...) might have a color of 0+0+6+2+5=13 mod 10 = 3, while 1/12 (0.083 33333...) might have a color of 0+0+8+3=11 mod 10 = 1.

Suppose you make an infinite list of decimals and color them with the same color. Then apply Cantor's diagonal method to produce an infinite list of new decimals which differ exactly in one digit and color each decimal in this new list a different color. And then keep repeating this process. (This was an informal arguement just to show the intuition behind the problem)

If we assign colors based on the last digit (the ones position) of the summation of all the digits of each number, it will not be possible for any two numbers to differ in only one digit and have the same color.