Very flavorful problem. By virtue of the triangle inequality, we want to find the probability that that longest of the three pieces doesn't exceed the combined length of the other two. Let X & Z denote the positions of the two random cuts (in metres) and let m & M be their minimum and maximum resp. I'll assume X & Z are independent. The probability we seek can be expressed as 1 - P(m > 1/2) - P(M < 1/2) - P(M - m > 1/2). Now,

- P(m > 1/2) = P(X > 1/2)*P(Z > 1/2) = (1/2)*(1/2) = 1/4
- P(M < 1/2) = P(X < 1/2)*P(Z < 1/2) = (1/2)*(1/2) = 1/4
- P(M - m > 1/2) = P(|X - Z| > 1/2) = (1/2)^3 + (1/2)^3 = 1/4 by geometric considerations (utilizing that (X,Z) is uniformly distributed on the unit square)

All in all, the desideratum is therefore 1/4.