Proposition: Each nonempty Polish space is the image of $\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$ under a continuous function.
(See for example Donald L. Cohn "Measure Theory" on page 250)
my Question: Is the constructed function $f$ even uniformly continuous?
In the proof he shows that $d(f(m),f(n)) \leq \frac{1}{k}$ for all $m,n \in \mathcal{N}$ with $m_i=n_i, i=1,...,k$. ($d$ denotes the metric on the Polish space).
Hence: Let $\epsilon >0$ and $N \in \mathbb{N}$ such that $\frac{1}{N} \leq \epsilon$. Choose $\delta:=\frac{1}{2^n}$. Then for $m,n \in \mathcal{N}$ with $d_{\mathcal{N}}(m,n)=\sum\limits_{k \in \mathbb{N}} \frac{1}{2^k} d_{discret}(m_k,n_k) < \delta=\frac{1}{2^N} \qquad$ ($d_{discret}$ is the discrete metric) it follows:
$d_{discret}(m_k,n_k) < 1$ for all $k=1,\dots,N$ and hence $m_k=n_k$ for $k=1,\dots,N$. So from the proof it follows that $d(f(m),f(n))\leq \frac{1}{N} < \epsilon$.
Your argument is correct. Perhaps it is a bit easier to see it by considering the following strongly equivalent metric on $\mathcal{N}$ given by $d(m,m):=0$ and if $m\neq n$, $$ d(m,n) := 2^{-\Delta(m,n)}, $$ where $\Delta(m,n)$ is the least index in which they differ.
By the strong equivalence, the notion of uniform continuity is the same for both metrics.