, 2012

The first step gives rise to one black square of area $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$. After the second step we obtain eight more squares of side $\frac{1}{9}$, the black region increasing thus by $\frac{8}{9^2}$. In the same manner, the third step increases the black area by $8^2 = 64$ black squares, each of area $\frac{1}{27^2}$, that is at this stage the black area becomes
$$\frac{1}{9} + \frac{8}{9^2} + \frac{8^2}{9^3}.$$
We conclude that after $1000$ steps, the area of the black region is
$$\begin{aligned} \frac{1}{9} + \frac{8}{9^2} + \frac{8^2}{9^3} + \dots + \frac{8^{999}}{9^{1000}} &= \frac{1}{9} \left( 1 + \frac{8}{9} + \left(\frac{8}{9}\right)^2 + \dots + \left(\frac{8}{9}\right)^{999} \right) \\ &= \frac{1}{9} \cdot \frac{1 - \left(\frac{8}{9}\right)^{1000}}{1 - \frac{8}{9}} = 1 - \left(\frac{8}{9}\right)^{1000}. \end{aligned}$$
It remains to prove that the last number is greater than $0.999$, i.e., $1 - \left(\frac{8}{9}\right)^{1000} > 0.999$, or equivalently $\left(\frac{8}{9}\right)^{1000} < 0.001$.

This easily follows by using a binomial expansion:
$$\left(\frac{9}{8}\right)^{1000} = \left(1 + \frac{1}{8}\right)^{1000} > \left(\frac{1000}{2}\right) \cdot \frac{1}{8^2} = \frac{1000 \cdot 999}{2 \cdot 64} > 1000.$$
This completes the proof.