Proof, [Proposition 8.2.7, Coh13]. Let $d: X^{2} \to [0, \infty)$ be a complete metric on $X$. To construct the desired map, it is sufficient to construct $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp}\subset 2^{X}$ such that:

1. (i)

   For each $N \in \natp$, $\bracsn{n_k}_{1}^{N} \subset \natp$, $C(n_{1}, \cdots, n_{N}) \subset X$ is closed and non-empty.

2. (ii)

   For each $N \in \natp$, $\bracsn{n_k}_{1}^{N} \subset \natp$, $\text{diam}(C(n_{1}, \cdots, n_{N})) \le 1/N$.

3. (iii)

   For each $N \in \natp$ and $\bracsn{n_k}_{1}^{N}\subset \natp$,

   \[C(n_{1}, \cdots, n_{N}) = \bigcup_{n_{N+1} \in \natp}C(n_{1}, \cdots, n_{N+1})\]

   
   
4. (iv)

   $X = \bigcup_{n_1 \in \natp}C(n_{1})$.

Since $X$ is Polish, there exists a countable dense subset $\seq{x_k}\subset X$. For each $n_{1} \in \natp$, let $C(n_{1}) = \ol{B(x_{n_1}, 1/2)}$, then $\bracsn{C(n_1)|n_1 \in \natp}$ satisfies (i) and (ii) by definition. As $\seq{x_k}$ is dense in $X$, $X = \bigcup_{n_1 \in \natp}C(n_{1})$, and (iv) is also satisfied.

Let $N \in \natp$ and suppose inductively that $\bracsn{C(n_1, \cdots, n_k)| \bracsn{n_j}_1^k \subset \natp,1 \le k \le N}$ has been constructed to satisfy (i)-(iv) for each $1 \le k \le N$.

Fix $\bracsn{n_k}_{1}^{N} \subset \natp$. By Proposition 8.1.2, there exists a countable dense subset $\seq{y_k}\subset C(n_{1}, \cdots, n_{N})$. For each $n_{N+1}\in \natp$, let $C(n_{1}, \cdots, n_{N+1}) = \ol{B(y_{n_{N+1}}, 1/[2(N+1)])}\cap C(n_{1}, \cdots, n_{N})$, then $\bracsn{C(n_1, \cdots, n_{N+1})|n_{N+1} \in \natp}$ satisfies (i) and (ii) by definition. Since $\seq{y_k}$ is dense in $C(n_{1}, \cdots, n_{N})$,

so (iii) is also satisfied.

Finally, suppose that $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp}\subset 2^{X}$ has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k}\in \mathscr{N}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})}= \bigcap_{N \in \natp}C(n_{1}, \cdots, n_{N})$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k}\in \mathscr{N}$ such that $x = f(\seq{n_k})$. As such, $f: \mathscr{N}\to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k}\in \mathscr{N}$ and $N \in \natp$ with $m_{k} = n_{k}$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 1/N$. Therefore $f \in C(\mathscr{N}; X)$.$\square$