Let $X$ be a nonempty Polish space. We construct a continuous onto map $f\colon \mathcal{N}\rightarrow X$ where $\mathcal{N}$ is Baire space.

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

1. $C()=X$
2. $C(n_1,\ldots,n_k)$ has diameter no more than $2^{-k}$
3. For any $k\ge 0$ and $n_1,\ldots n_k\in\mathbb{N}$ then \begin{equation}\label{eq:1} C(n_1,\ldots,n_k)=\bigcup_{m=1}^\infty C(n_1,\ldots,n_k,m). \end{equation}

We now define the function $f\colon\mathcal{N}\rightarrow X$ For any $n\in\mathbb{N}$ choose a sequence $x_k\in C(n_1,\ldots,n_k)$ 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 and has a limit $x$ As the sets $C(n_1,\ldots,n_k)$ are closed, they contain $x$ and, \begin{equation}\label{eq:2} \bigcap_{k=1}^\infty C(n_1,\ldots,n_k)\not=\emptyset. \end{equation}In fact, this has diameter zero, and must contain a single element, which we define to be $f(n)$

This defines the function $f\colon\mathcal{N}\rightarrow X$ We show that it is continuous. If $m,n\in\mathcal{N}$ satisfy $m_j=n_j$ for $j\le k$ then $f(m),f(n)$ are in $C(m_1,\ldots,m_k)$ 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 () allows us to choose $n_1,n_2,\ldots$ such that $x\in C(n_1,\ldots,n_k)$ for all $k\ge 0$ If $n=(n_1,n_2,\ldots)$ then $x$ and $f(n)$ are both in the set in equation () which, since it is a singleton, gives $f(n)=x$ Hence, $f$ is onto.