proof of Baire space is universal for Polish spaces

Let $X$ be a nonempty Polish space. We construct a continuous onto map $f:𝒩\to X$, where $𝒩$ is Baire space.

Let $d$ be a complete metric on $X$.We choose nonempty closed subsets $C(n_1,\mathrm{\dots },n_k)\subseteq X$ for all integers $k\ge 0$ and $n_1,\mathrm{\dots },n_k\in \mathbb{N}$ satisfying the following.

1. $C()=X$.

2. $C(n_1,\mathrm{\dots },n_k)$ has diameter no more than $2^{-k}$.

3. For any $k\ge 0$ and $n_1,\mathrm{\dots }{n}_k\in \mathbb{N}$ then

   $$C(n_1,\mathrm{\dots },n_k)=\bigcup _{m=1}^{\mathrm{\infty }}C(n_1,\mathrm{\dots },n_k,m).$$(1)

This can be done by induction. Suppose that the set $S=C(n_1,\mathrm{\dots },n_k)$ has already been chosen for some $k\ge 0$. As $X$ is separable, $S$ can be covered by a sequence of closed sets $S_1,S_2,\mathrm{\dots }$. Replacing $S_j$ by $S_j\cap S$, we suppose that $S_j\subseteq S$. Then, remove any empty sets from the sequence. If the resulting sequence is finite, then it can be extended to an infinite sequence by repeating the last term.We can then set $C(n_1,\mathrm{\dots },n_k,n_{k+1})=S_{n_{k+1}}$.

We now define the function $f:𝒩\to X$. For any $n\in \mathbb{N}$ choose a sequence $x_k\in C(n_1,\mathrm{\dots },n_k)$. Since this has diameter no more than $2^{-k}$ it follows that $d(x_j,x_k)\le 2^{-k}$ for $j\ge k$. So, the sequence is Cauchy (http://planetmath.org/CauchySequence) and has a limit $x$. As the sets $C(n_1,\mathrm{\dots },n_k)$ are closed, they contain $x$ and,

$$\bigcap _{k=1}^{\mathrm{\infty }}C(n_1,\mathrm{\dots },n_k)\ne \mathrm{\varnothing }.$$(2)

In fact, this has diameter zero, and must contain a single element, which we define to be $f(n)$.

This defines the function $f:𝒩\to X$. We show that it is continuous. If $m,n\in 𝒩$ satisfy $m_j=n_j$ for $j\le k$ then $f(m),f(n)$ are in $C(m_1,\mathrm{\dots },m_k)$ which, having diameter no more than $2^{-k}$, gives $d(f(m),f(n))\le 2^{-k}$. So, $f$ is indeed continuous.

Finally, choose any $x\in X$. Then $x\in C()$ and equation (1) allows us to choose $n_1,n_2,\mathrm{\dots }$ such that $x\in C(n_1,\mathrm{\dots },n_k)$ for all $k\ge 0$. If $n=(n_1,n_2,\mathrm{\dots })$ then $x$ and $f(n)$ are both in the set in equation (2) which, since it is a singleton, gives $f(n)=x$. Hence, $f$ is onto.