Does a real number with this decimal expansion for $r$ and $r^2$ exist? Does  there  exist  a  real  number $0< x <1$,  such  that the  decimal expansions  of $x$ and $x^2$
are  the  same,  starting  from  the
millionth term, and neither expansion has an infinite tail of zeroes?
I was thinking $x=0.\overline{999}$, but does that work? Isn't that just equal to 1 which is not allowed.? If this works, how would I prove it?

We can concoct an example quite easily. Suppose we want the difference between $x$ and $x^2$ to be 0.1:
$$x-x^2=0.1$$
where the order $x-x^2$ is mandated by $0<x<1$, so $x^2<x$. Solving this, we get two admissible values $x=\frac{1\pm\sqrt{0.6}}2$.
Thus (taking $x=\frac{1+\sqrt{0.6}}2$) we have
$$x=0.88729833\dots$$
$$x^2=0.78729833\dots$$
so their decimal expansions agree after the first place, and indeed after the millionth place.
Any number $0<k<0.25$ with a terminating decimal expansion such that $\sqrt{1-4k}$ does not terminate can be used in place of the 0.1 in $x-x^2=0.1$.
