Probability of forming triangle from breaking a stick

The lengths of the sides of the stick will be $1-A, A-B$, and $B$. For the stick to form a valid triangle, the following three conditions must hold by the triangle inequality:

$$\begin{align*} 1-A + A-B>B \to B&<\frac{1}{2} \\ 1-A + B>A-B \to B&>A-\frac{1}{2} \\ A-B+B > 1-A \to A&>\frac{1}{2} \end{align*}$$

You did not include the first condition, $B<1/2$, which threw off your answer. Other than this, everything else was solved correctly; your answer would have been correct if only the second and third conditions were needed.