Theorem 1

A subspace $Y$ of a Polish space $X$ is Polish if and only if $Y$ is the intersection of a countable family of open sets in $X$.

The sufficiency of the condition follows immediately from Propositions 2 and 3. To show necessity, let $d$ be a metric compatible with the topology of $Y$ and with respect to which $Y$ is complete. Let $\overline{Y}$ be the closure of $Y$ in $X$. For each integer $n > 0$, let $Y_n$ be the set of all $x \in \overline{Y}$ which have an open neighbourhood $U$ such that the diameter of $U \cap Y$ (with respect to the metric $d$) is $\leq 1/n$. Clearly $Y_n$ is open in $\overline{Y}$ and contains $Y$. Let $x \in \bigcap_n Y_n$; then $x \in \overline{Y}$, and the trace on $Y$ of the neighbourhood filter of $x$ in $X$ is a Cauchy filter (with respect to $d$); hence this filter converges to a point of $Y$, and thus $x \in Y$. Hence $Y = \bigcap_n Y_n$.

For each $n$, let $H_n$ be an open subset of $X$ such that $H_n \cap \overline{Y} = Y_n$, and let $(U_m)$ be a sequence of open subsets of $X$ such that $\overline{Y} = \bigcap_m U_m$ (§ 2, no. 5, Proposition 7); then $Y$ is the intersection of the countable family of open sets $(H_n \cap U_m)$.