Let $U=\{1,2,3,4,5,6\}$ and S consists of $S_1=\{1,2,3,4\}$, $S_2=\{1,2,5\}$, $S_3=\{3,4,6\}$. The algorithm would give an answer of $S_1$, $S_2$, $S_3$ but the correct answer is $S_2$,$S_3$ because it has the fewest subsets and covers $U$.