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 $$\begin{array}{rl}X+(Y-X)& >1-Y\\ (Y-X)+(1-Y)& >X\\ X+(1-Y)& >Y-X\end{array}$$ 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 $X\le 1$ and $Y\le 1$. Hence the required probability is $2\times 1/8=1/4$.