A set in the Borel $\sigma$-algebra over $[0,1]$ that isn't in the algebra generated by open sets

The Borel algebra on $[0,1]$ is by defintion a $\sigma$-algebra, the smallest one containing every open subset of $[0,1]$. I'm wondering how the Borel algebra differs from the algebra generated by the open subsets of $[0,1]$.

What's an example of a set in the Borel algebra that you can't obtain by closing open subsets of $[0,1]$ under complements and finite unions?

Added. Does any countable dense set work? Obviously we have to choose a Borel set that is neither open nor closed. The set of rationals in $[0,1]$ fits the bill here, and my intuition is that it cannot be written using finitely many set operations with open sets, but I'm not quite sure how to prove it.

Solution 1:

Claim: For any set $X$ obtainable the boundary is nowhere dense.

Proof: Boundary of open set is nowhere dense. Now for the operations: 1) complement - since boundary is the same for a set and its complement, this is automatic 2) union - boundary of a union is subset of union of boundaries, and so it is a subset of a union of two nowhere dense sets, so is a subset of a nowhere dense set, and so is nowhere dense.

Thus we can not obtain the rationals, since their boundary is all of $[0,1]$. The same is true for all countable dense sets.