Prove that $\lfloor0.999\dots\rfloor= ?$ $0$ or $1$?

Your first assertion is correct. The other observation just says that the function $x\mapsto\lfloor x\rfloor$ is not continuous.

For one thing, $0.999...$ is exactly equal to $1$.

To prove that $0.999... = 1$, we use sums of infinite geometric sequences. We know that $$\sum_{k=0}^\infty r^k = \frac{1}{1-r} \forall \;\left|r\right|\lt1$$ It is fairly simple to prove this statment, although I won't go into that. For our specific instance, we have $$0.999... = \frac{9}{10}\sum_{k=0}^\infty (\frac{1}{10})^k = \frac{9}{10} \cdot\frac{1}{1-\frac{1}{10}} = \frac{9}{10}\cdot\frac{10}{9}=1$$

Using the substitution property, we can substitute any alternate representation of a number into an equation, and it will yield the same result. Therefore,

$$\lfloor0.999...\rfloor = \lfloor1\rfloor = 1$$