I am going to answer the question that you all ask whenever you see a metric space "is it complete?"
$\mathbf{Proposition}$ The metric $d'$ on $\mathbb{N}^{\mathbb{N}}$ is not a complete metric.
$\mathbf{Proof}$ Let $f_{n}:\mathbb{N}\rightarrow\mathbb{N}$ denote the function where $f_{n}(n)=1$ and $f_{n}(m)=1$ whenever $m\neq n$. I claim that the sequence $(f_{n})_{n}$ is a Cauchy sequence that does not converge.
To see that the sequence $(f_{n})_{n}$ is Cauchy, we take note that for all $N$, if $n,m>N$, then the functions $f_{n},f_{m}$ both have the same prefixes of length $n$ and the same infixes of length at most $n$, so $d'(f_{n},f_{m})\leq\frac{1}{N}$. We therefore conclude that the sequence $(f_{n})_{n}$ is Cauchy.
On the other hand, the sequence $(f_{n})_{n}$ does not converge to any point in $\mathbb{N}^{\mathbb{N}}$ with respect to $d'$. If $f_{n}\rightarrow f$ with respect to $d'$, then for all $r\in\mathbb{N}$, we have $f_{n}(r)\rightarrow f(r)$. This shows that $f(r)=0$ for all $r$. However, $f_{n}$ does not converge to $f$ with respect to $d'$ since $f_{n}$ contains the infix $1$ where $f$ does not. $\mathbf{QED}$
Since $d'$ is not a complete metric, it would probably be a good idea to look at the completion of $(\mathbb{N}^{\mathbb{N}},d')$ and not simply $(\mathbb{N}^{\mathbb{N}},d')$.