Set of closed intervals $[x, y] \subset [0, 1]$ $\pi$-system which generates Borel $\sigma$-algebra on $[0, 1]$? How do I see that the set of closed intervals $[x, y] \subset [0, 1]$ is a $\pi$-system which generates the Borel $\sigma$-algebra on $[0, 1]$?

To see this, first verify that the set of closed intervals is a $\pi-$system which is the result of having finite intersection of closed set as closed.
To see that it generates Borel $\sigma-$algebra, it is enough to show that every open set $(x,y)$ (and $(x,1]$ and $[0,x)$ belongs to $\sigma-$algebra of closed set. $(x,1]$ and $[0,x)$ trivially belong to $\sigma-$algebra, since they are complements of closed sets. On the other hand, countable union of $[x-\frac 1n,y-\frac 1n]$ over $n\in\mathbb N$ yields the open set $(x,y)$. So all open sets are  in $\sigma-$algebra of closed sets and therefore it includes Borel $\sigma-$algebra. The reverse direction can be shown by seeing that each closed set is already in Borel $\sigma-$algebra.
