Assume that we break the stick at points $X$ and $Y$. Assume $X < Y$. Then for the stick to form a triangle, the three lengths $X, Y-X$ and $1-Y$ should satisfy the following three inequalities \begin{align} X+(Y-X) &> 1-Y\newline (Y-X) + (1-Y) &> X\newline X + (1-Y) &> Y-X \end{align} which is nothing but the triangluar region between the points $(0, 0.5), (0.5, 0.5)$ and $(0.5, 1)$ and has the area of $1/8$. We should also consider the case $Y < X$ and by symmetry, the area is same. Now, $X$ and $Y$ comprise of the entire square region $X \leq 1$ and $Y \leq 1$. Hence the required probability is $2 \times 1/8 = 1/4$.