$\mathcal{N}:=\mathbb{N}^{\mathbb{N}}$
Let $d(m,n)=\sum\limits_{n=1}^{\infty}\:\frac{1}{2^n}\: d_n(m_n,n_n)$ be the product metric on $\mathcal{N}$, where $d_n$ denotes the discrete metric. I want to show that $d(m,n)\leq \frac{1}{2^k}$ for $m=(m_1,m_2,\dots),\: n=(n_1,n_2,\dots) \in \mathcal{N}$ with $m_n=n_n$ for all $n=1,\dots,k$:
\begin{align}d(m,n)&=\sum_{n=1}^{\infty}\:\frac{1}{2^n}\: d_n(m_n,n_n)=\sum_{n=k+1}^{\infty} \frac{1}{2^n}\, d_n(m_n,n_n) \leq \sum_{n=k+1}^{\infty}\frac{1}{2^n}\\ & = \sum\limits_{n=0}^{\infty}\:\frac{1}{2^n}-\sum_{n=0}^{k}\frac{1}{2^n} = 2-\frac{1-\cfrac{1}{2^{k+1}}}{1-\cfrac{1}{2}} =\frac{1}{2^k} \end{align}
Is every step well defined?
Yes it is correct but maybe some justifications are needed. The first equality is the definition of $d$; the second one is due to the fact that $m_n=n_n$ for all $n\in\{1,\dots,k\}$ hence $d_n(m_n,n_n)=0$ for these $n$ and the inequality due to the fact that $d_n$ is always smaller than one.