What is the probability of two out of three events happening?

Your approach works, and your answer is correct.

It can also be calculated straight forwards as:

$\begin{align} \mathsf P(AB\cup AC\cup BC) & = \mathsf P(AB)+\mathsf P(A^cBC)+\mathsf P(AB^cC) \\[1ex] & = \mathsf P(A)\mathsf P(B) + \bigg(\mathsf P(A^c)\mathsf P(B)+\mathsf P(A)\mathsf P(B^c)\bigg)\mathsf P(C) \\[0ex] &= \frac{9^2}{10^2} + \bigg(\frac{1\cdot 9}{10^2} + \frac{9\cdot 1}{10^2}\bigg)\frac 6{ 10} \\[1ex] & = \frac{918}{1000} \end{align}$

Can confirm that is the right answer, calculated it directly and obtained the same number.