What is $\left\lfloor0.\overline{9}\right\rfloor$? We know that $0.\overline{9} = 1$ but then what is $\left\lfloor0.\overline{9}\right\rfloor$?

My thought process went:
$0.\overline{9} = 1$ so therefore $\left\lfloor0.\overline{9}\right\rfloor = \left\lfloor1\right\rfloor$ 

But also $\left\lfloor0.9\right\rfloor = 0$ and it shouldn't change no matter how many numbers you add on to the back of it.

So what is the right answer, if there is one, and why?
As you say, $0.\overline9=1$. Thus, $0.\overline9$, like $2-1$, $\frac33$, etc., is just a different notation for the number $1$, and consequently $\left\lfloor0.\overline9\right\rfloor=\lfloor1\rfloor=1$.

Added: The fact that $\lfloor 0.9\ldots9\rfloor=0$ for any finite string of $9$s is irrelevant: the floor function is not continuous from the left at integers. In effect you’re saying that $\lfloor 1\rfloor$ ought to be $0$ simply because $\left\lfloor 1-10^{-n}\right\rfloor=0$ for each $n\in\Bbb Z^+$.
To me the intuitive explanation is that if you put any number of 9s after the 0, $\lfloor 0.999...99 \rfloor =0$. But $0.\overline{9}$ has infinite 9s after the 0, and infinity isn't a number. So the pattern doesn't hold.

Cameron Williams pointed out the rigorous answer, which is that you're implicitly assuming that because a sequence of numbers (0.9,0.99,0.999...) gets closer to 1, then if you apply a function to them (in this case, $\lfloor x \rfloor$), the the resulting sequence ($\lfloor 0.9\rfloor$,$\lfloor 0.99\rfloor$,$\lfloor 0.999\rfloor$...) should also get closer to  $\lfloor 1\rfloor$. But functions don't always do this. If a certain function does have this property, we say that the function is continuous at that point, but $\lfloor x\rfloor$ is not continuous at 1. In fact, you've more or less given a proof that $\lfloor x\rfloor$ is discontinuous at 1.