Every Polish space has a countable $\pi$-system

Question

The $\sigma$-algebra of Borel $\mathcal{B}([0,1])$ has a countable $\pi$-system $\mathcal D$ such that $\sigma(\mathcal{D})=\mathcal{B}([0,1])$, namely, it suffices to take

$$ \mathcal{D}:=\{(a,b),(a,b],[a,b),[a,b]\colon a,b\in \mathbb{Q}\cap [0,1]\} $$

Suppose that $E$ is a Polish space. Is it true that there exists a countable $\pi$-system $\mathcal{D}\subset\mathcal{B}(E)$ such that $\sigma(\mathcal{D})=\mathcal{B}(E)$?

Answer 1

Any Polish space is second countable. If $(U_n)$ is a countable base then finite intersections of $U_n$'s gives a $\pi-$ system which generates the Borel sigma algebra.