Proposition 2

Every open subspace of a Polish space is Polish.

Let $X$ be a Polish space, let $d$ be a metric on $X$ compatible with its topology, and let $U \neq X$ be an open subset of $X$. Let $V$ be the subset of $\mathbf{R} \times X$ consisting of all points $(t, x)$ such that

$V$ is closed by Proposition 3 of § 2, no. 2 and is therefore Polish (Proposition 1). Since the restriction to $V$ of the projection $\mathrm{pr}_2 : \mathbf{R} \setminus X \to X$ is a homeomorphism of $V$ onto $U$ ($§ 2,$ no. 2, Proposition 3), $U$ is a Polish subspace of $X$.