$0.999\dots$ is the decimal expansion where every digit after the decimal point is a $9$. By definition, it is the value of the series ${\sum }_{k=1}^{\infty }9\cdot {10}^{-k}$. This value is in turn defined as the limit of the sequence $({\sum }_{k=1}^{n}9\cdot {10}^{-k}{)}_{n\in N}$. Let $a_n$ denote the $n$th term of this sequence. I claim the limit is $1$. To prove this, we have to show that for any $\epsilon >0$, there is some $N\in N$ such that for every $n>N$, $|1-a_n|<\epsilon$.

Let's prove by induction that $1-a_n={10}^{-n}$. Since $a_0$ is the sum of {$0$ terms, $a_0=0$, so $1-a_0=1={10}^0$. If $1-a_i={10}^{-i}$, then

$$\begin{array}{cc}1-a_{i+1}& =1-(a_i+9\cdot {10}^{-(i+1)})\\ & =1-a_i-9\cdot {10}^{-(i+1)}\\ & ={10}^{-i}-9\cdot {10}^{-(i+1)}\\ & =10\cdot {10}^{-(i+1)}-9\cdot {10}^{-(i+1)}\\ & ={10}^{-(i+1)}\end{array}$$

So $1-a_n={10}^{-n}$ for all $n$. What remains to be shown is that ${10}^{-n}$ eventually gets (and stays) arbitrarily small; this is true by the Archimedean_property and because ${10}^{-n}$ is monotonically decreasing.

Decimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there's no reason different decimal expansions have to represent different real numbers.

Decimal expansions go on infinitely, but no farther. $0.000\dots 001$ doesn't represent a real number because the $1$ is supposed to be after infinitely many $0$s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $0.000\dots 001$ to represent, it would be $0$.

The sequence gets arbitrarily close to $1$, so its limit is $1$. It doesn't matter that all of the terms are less than $1$.

There are infinitely many $9$s in $0.999\dots$, so when you shift it over a digit there are still the same amount. And the "decimal expansion" $8.999\dots 991$ doesn't make sense, because it has infinitely many digits and then a $1$.