Is there a countable collection $(E\_n)\_{n \in \mathbf{N}}$ of Borel subsets of $I = [0,1]$ such that, for every Borel subset $E$ of $I$ and every $\epsilon > 0$ there exists $n,m$ with $E\_n \subset E \subset E\_m$ and $\mu(E\_m \setminus E\_n) < \epsilon$ where $\mu$ denotes the Lebesgue measure ?

Equivalently : is there a countable collection $(E\_n)$ of Borel subsets such that for every Borel $E$ and $\epsilon>0$ there exists $n$ such that $E \subset E\_n$ and $\mu(E\_n \setminus E) < \epsilon$ ?

Equivalently, by regularity of the Lebesgue measure: is there a countable collection $(E\_n)$ of Borel subsets such that for all open subset $E$ and $\epsilon>0 $ there exists $n$ such that $E \subset E\_n$ and $\mu(E\_n \setminus E) < \epsilon\ $?

Equivalently, again by regularity of the Lebesgue measure : is there a countable collection $(E\_n)$ of open subsets such that for all open subset $E$ and $\epsilon>0 $ there exists $n$ such that $E \subset E\_n$ and $\mu(E\_n \setminus E) < \epsilon\ $?

I suspect that the answer is `no', but I cannot find an argument.