Proof. (1): Let $X$ be a Polish space with complete metric $d$ and $A \subset X$ be a closed subset, then $A$ is second countable. By Proposition 6.7.3, $A$ is complete with respect to $d$.

(2): Let $X$ be a Polish space with complete metric $d$ and $U \subset X$ be open. Assume without loss of generality that $U \subsetneq X$ and $d(X \times X) \subset [0, 1]$. Define

then $d_{U}$ is a metric on $U$. Since $d \le d_{U}$, the topology induced by $d_{U}$ is finer than the topology induced by $d$. On the other hand, the mapping $x \mapsto d(x, U^{c})$ is continuous, so the topology induced by $d_{U}$ is coarser than the topology induced by $d$. Therefore $d_{U}$ induces the subspace topology of $U$.

Now, let $\seq{x_n}\subset U$ be a Cauchy sequence with respect to $d_{U}$, then there exists $N \in \natp$ such that $d_{U}(x_{m}, x_{n}) \le 1$ for all $m, n \ge N$. Thus

and $\delta = \inf_{n \in \natp}d(x_{n}, U^{c}) > 0$. Since $d \le d_{U}$, $\seq{x_n}$ is Cauchy with respect to $d$ as well. Thus as $\bracsn{x \in X|d(x, U^c) \ge \delta}\subset U$ is a closed subset of $X$, there exists $x \in U$ such that $x_{n} \to x$ as $n \to \infty$. Therefore $U$ is complete with respect to $d_{U}$.

(3): By Proposition 5.5.7, Proposition 6.7.2, and Theorem 6.3.10.$\square$