Corollary 1

A space $X$ is Polish if and only if it is homeomorphic to a countable intersection of open sets in the cube $\mathbf{I}^{\mathbf{N}}$, where $\mathbf{I}$ is the interval $[0, 1]$ of $\mathbf{R}$.

The condition is clearly sufficient, and it is necessary because every metrizable space of countable type is homeomorphic to a subspace of $\mathbf{I}^{\mathbf{N}}$ (§ 2, no. 8, Proposition 12).