(I haven't done real math since forever)

Part 1: Proving If x,y are distinct positive real numbers and x < y, there is a z in R such that x < z < y.

Let y = x + e, where e is in R. e is also positive because:

x < y
x < x + e
0 < e

x < z < y becomes x < z < x + e. We can then choose z = x + 0.5e so that

x < z < y
x < x + 0.5e < x + e
0 < 0.5e < e
0.5e - 0.5e < 0.5e + 0 < 0.5e + 0.5e
-0.5e < 0 < 0.5e

Since we've shown earlier that e > 0, -0.5e < 0 < 0.5e is true no matter what e.

Part 2: Decimal Notation

Decimal notation is written as the sequence a~m~a~m-1~...a~1~a~0~.b~1~b~2~...b~n~, where a,b are in the set {0,1,2,3,4,5,6,7,8,9}. The sequence a~m~a~m-1~...a~1~a~0~.b~1~b~2~...b~n~ itself is of the number a~m~10^m^ + a~m-1~10^m-1^ + ... + a~0~10^0^ + b~1~10^-1^ + b~2~10^-2^ + ... + b~n~10^-n^.

Part 3: The actual proof through proof by contradiction

Let x = 0.999... and y = 1. This means there exists a z that is between 0.999... and 1. So let's construct z = a~0~.b~1~b~2~...b~n~. a~0~ has to be either a 0 or 1 and since there is no number smaller than 1 with 1 as its first digit, a~0~=0. z = 0.b~1~b~2~...b~n~

0.999... = 9*10^-1^ + 9*10^-2^ + 9*10^-3^ + ... while z = b~1~10^-1^ + b~2~10^-2^ + ... + b~n~10^-n^. If 0.999... < z, then

0.999... - z < 0
9*10^-1^ + 9*10^-2^ + 9*10^-3^ + ... - (b~1~10^-1^ + b~2~10^-2^ + ... + b~n~10^-n^) < 0
(9*10^-1^ - b~1~10^-1^) + (9*10^-2^ - b~2~10^-2^) + (9*10^-3^ - b~3~10^-3^) + ... < 0
(9 - b~1~)*10^-1^ + (9 - b~2~)*10^-2^ + (9 - b~3~)*10^-3^ + ... < 0

But as we've established in Part 2, b~1~, b~2~, b~3~, etc have to be from the set {0,1,2,3,4,5,6,7,8,9}, meaning (9 - b~n~) > 0 for any n. Therefore, (9 - b~1~)*10^-1^ + (9 - b~2~)*10^-2^ + (9 - b~3~)*10^-3^ + ... cannot be less than 0.

The theorem requires for x and y to be distinct positive real numbers and x < y, and since 0.999... and 1 can be trivially shown to be positive real numbers, this means that 0.999... and 1 are not distinct. In other words, 0.999... = 1.