I have proved the following result in arXiv:1005.3371v8 Lemma 3.31:
Let $n$ be a natural number greater than or equal to 2. There exists a bijection $f$ from $\mathbb{N}$ onto $\mathbb{Z}^n$ so that $\Vert f(k + 1) - f(k) \Vert_\infty = 1$ for all natural numbers $k$.
Here is a simpler proof:
$$ \newcommand{\naturalnumbers}{\mathbb{N}} \newcommand{\integers}{\mathbb{Z}} \newcommand{\positiveintegers}{{\mathbb{Z}_+}} \newcommand{\natnumberset}[1]{N(#1)} \newcommand{\integerset}[1]{Z(#1)} \newcommand{\setzeroton}[1]{Z_{0+}(#1)} \newcommand{\zeron}[1]{0_{#1}} \newcommand{\card}[1]{\#{#1}} \newcommand{\onto}{\twoheadrightarrow} \newcommand{\setsep}{:} \newcommand{\spaceafter}{\;\;\;\;} $$
$$ \textbf{Definitions} $$
$$\natnumberset{k} := \{ j \in \naturalnumbers \setsep j \leq k \}, \spaceafter k \in \naturalnumbers$$
$$\integerset{k} := \{ j \in \integers \setsep \vert j \vert \leq k \}, \spaceafter k \in \naturalnumbers$$
$$\zeron{n} := (0,\ldots,0) \in \integers^n, \spaceafter n \in \positiveintegers$$
$$A_{n,k} := \left( \integerset{k} \right)^n, \spaceafter n \in \positiveintegers, \; k \in \naturalnumbers$$
$$A_{n,-1} := \emptyset$$
$$B_{n,k} := A_{n,k} \setminus A_{n,k-1}, \spaceafter n \in \positiveintegers, \; k \in \positiveintegers$$
$$B_{n,0} := A_{n,0} = \{ \zeron{n} \}, \spaceafter n \in \positiveintegers$$
Let $S \subset \naturalnumbers$, $n \in \positiveintegers$, and $f$ a function from $S$ into $\integers^n$. We say that $f$ $\textit{preserves neighbours}$ if $k, k+1 \in S$ implies $\Vert f(k+1)-f(k) \Vert_\infty = 1$.
Define $$\beta_{2,k}(i) := \left\{ \begin{array}{ll} (-k+1+c,-k) ; & i = c \in \natnumberset{2k-1} \\ (k, -k+c+1) ; & i = 2k + c, c \in \natnumberset{2k-1} \\ (k-c-1, k) ; & i = 4k + c, c \in \natnumberset{2k-1} \\ (-k, k-c-1) ; & i = 6k + c, c \in \natnumberset{2k-1} \end{array} \right.$$ where $i \in \natnumberset{8k-1}$ and $k \in \positiveintegers$. Define also $\beta_{2,0}(0) := (0,0)$.
Define $ \alpha_{2,k}(i) := \beta_{2,j}(i-\card{A_{2,j-1}}) $ where $\card{A_{2,j-1}} \leq i < \card{A_{2,j}}$ and $i \in \natnumberset{\card{A_{2,k}}-1}$, $j \in \natnumberset{k}$ and $k \in \positiveintegers$. When $k=0$ let $\alpha_{2,0}(0) = (0,0)$. Let also $\alpha'_{2,k}(i) := \alpha_{2,k}(\card{A_{2,k}}-1-i)$ for $i \in \natnumberset{\card{A_{2,k}}-1}$.
If $k < j$ we define $\sum_{i=j}^k x_i = 0$.
$$ \textbf{Lemma 1} $$
Let $k \in \naturalnumbers$. Function $\beta_{2,k}$ is a bijection from $\natnumberset{\card{B_{2,k}}-1}$ onto $B_{2,k}$ preserving neighbours and $\alpha_{2,k}$ is a bijection from $\natnumberset{\card{A_{2,k}}-1}$ onto $A_{2,k}$ preserving neighbours.
$$ \textbf{Lemma 2} $$
Let $n \in \naturalnumbers$, $n \geq 3$. There exist bijections $\beta_{n,k} : \natnumberset{\card{B_{n,k}}-1} \onto B_{n,k}$, $k \in \naturalnumbers$, so that $\beta_{n,k}$ preserves neighbours, $\beta_{n,k}(0) = (k, \zeron{n-1})$, and $\beta_{n,k}(\card{B_{n,k}}-1) = (-k, \zeron{n-1})$ for all $k \in \naturalnumbers$.
$$ \textit{Proof.} $$ We prove case (n = 3) first. Suppose that $k \in \positiveintegers$. Let $$C_0 := \{(k,\mathbf{x}) \setsep \mathbf{x} \in A_{2,k}\}$$ and $$p_0(i) := (k,\alpha_{2,k}(i))$$ for $i \in \natnumberset{\card{A_{2,k}}-1}$. Let $$C_m := \{(k-m,\mathbf{x}) \setsep \mathbf{x} \in B_{2,k}\}$$ and $$p_m(i) := (k-m,\beta_{2,k}(i))$$ for $i \in \natnumberset{\card{B_{2,k}}-1}$ and $m = 1, \ldots, 2k-1$. Let $$C_{2k} := \{(-k,\mathbf{x}) \setsep \mathbf{x} \in A_{2,k}\}$$ and $$p_{2k}(i) := (-k,\alpha'_{2,k}(i))$$ for $i \in \natnumberset{\card{A_{2,k}}-1}$. Define $$\beta_{3,k}(i) := p_m\left(i - \sum_{m'=0}^{m-1}\card{C_{m'}}\right)$$ when $$ \sum_{m'=0}^{m-1} \card{C_{m'}} \leq i < \sum_{m'=0}^m \card{C_{m'}} $$ and $i \in \natnumberset{\card{B_{3,k}}-1}$ Now functions $\beta_{3,k}$ satisfy the conditions of the lemma.
Suppose that the lemma is true for some $n \in \naturalnumbers$, $n \geq 3$. Define $\beta^1_{n,k} := \beta_{n,k}$ for all $k \in \naturalnumbers$, $\beta^{-1}_{n,k}(i) := \beta_{n,k}(\card{B_{n,k}}-1-i)$ for all $i \in \natnumberset{\card{B_{n,k}}-1}$, $k \in \naturalnumbers$. Define also $ \alpha_{n,k}(i) := \beta^{(-1)^{k-j}}_{n,j}(i-\card{A_{n,j-1}}) $ where $\card{A_{n,j-1}} \leq i < \card{A_{n,j}}$ and $i \in \natnumberset{\card{A_{n,k}}-1}$, $j \in \natnumberset{k}$ and $k \in \positiveintegers$. When $k=0$ let $\alpha_{n,0}(0) = \zeron{n}$. Let also $$ \alpha'_{n,k}(i) := \beta^{(-1)^j}_{n,k-j}\left( i - \sum_{j'=0}^{j-1} \card{B_{n,k-j'}} \right) $$ for $i \in \natnumberset{\card{A_{n,k}}-1}$ where $$ \sum_{j'=0}^{j-1} \card{B_{n,k-j'}} \leq i < \sum_{j'=0}^j \card{B_{n,k-j'}} . $$ Let $k \in \positiveintegers$. Let $$C_0 := \{(k,\mathbf{x}) \setsep \mathbf{x} \in A_{n,k}\}$$ and $$p_0(i) := (k,\alpha_{n,k}(i))$$ for $i \in \natnumberset{\card{A_{n,k}}-1}$. Let $$C_m := \{(k-m,\mathbf{x}) \setsep \mathbf{x} \in B_{n,k}\}$$ and $$ p_m(i) := (k-m, \beta^{(-1)^m}_{n,k}(i)) $$ for $i \in \natnumberset{\card{B_{n,k}}-1}$ and $m = 1, \ldots, 2k-1$. Let $$C_{2k} := \{(-k,\mathbf{x}) \setsep \mathbf{x} \in A_{n,k}\}$$ and $$p_{2k}(i) := (-k,\alpha'_{n,k}(i))$$ for $i \in \natnumberset{\card{A_{n,k}}-1}$. Define $$\beta_{n+1,k}(i) := p_m\left(i - \sum_{m'=0}^{m-1}\card{C_{m'}}\right)$$ when $$ \sum_{m'=0}^{m-1} \card{C_{m'}} \leq i < \sum_{m'=0}^m \card{C_{m'}} . $$ and $i \in \natnumberset{\card{B_{n+1,k}}-1}$ Now functions $\beta_{n+1,k}$ satisfy the conditions of the lemma.
$$ \textbf{Theorem} $$ Let $n \in \naturalnumbers$, $n \geq 2$. There exists a bijection $\sigma_n : \naturalnumbers \onto \integers^n$ so that $\sigma_n$ preserves neighbours.
$$ \textit{Proof.} $$ Case $n = 2$ is simple and we don't prove it here. Case $n \geq 3$ follows from lemma 2 when we set $$ \sigma_n(i) := \beta^{(-1)^k}_{n,k}\left( i - \card{A_{n,k-1}}\right) $$ for $$ \card{A_{n,k-1}} \leq i < \card{A_{n,k}} $$ and $i \in \naturalnumbers$.
Could somebody verify the proof?
Tommi Höynälänmaa