Is there a map $s: \mathbb{Q}\times\mathbb{Q} \to \mathbb{N}$ with the following properties?
For all $z\in\mathbb{Z}$, the restriction $s|_{[z,z+1)\times [z,z+1)}: [z,z+1)\times [z,z+1) \to \mathbb{N}$ is bijective, and
For all $q\in\mathbb{Q}$, the restrictions $s|_{\{q\} \times \mathbb{Q}}: \{q\} \times \mathbb{Q} \to \mathbb{N}$ and $s|_{\mathbb{Q} \times \{q\}}: \mathbb{Q} \times \{q\} \to \mathbb{N}$ are bijective.
All sets involved are countably infinite, so there is a "step-by-step" construction which achieves the desired result.
Specifically, let us consider the following collection of conditions:
We can enumerate this set of all conditions as a single list $P_i,i\in\Bbb N$, and our goal is to successively assign values to points in $\Bbb Q^2$ to satisfy all conditions them all while making sure that the restrictions to each square or line is injective. The key is that at every step only finitely many values will have been assigned, leaving us with infinitely many possibilities to satisfy a condition and not break injectivity.
If $P_i$ is a condition of the form $A(z,n),B(q,n)$ or $C(q,n)$ which has not been satisfied yet, then we can pick some point in the relevant square or line which is not in any of the other squares of lines to which a value has been assigned yet, and give it value $n$. If $P_i$ is of the form $D(q,r)$, then we can assign to $(q,r)$ a value different from any values so far assigned to points on the square or lines which $(q,r)$ lies on.
We can do this with permutations.
Let $ \def\N{\mathbb{N}} \def\Q{\mathbb{Q}} \def\Z{\mathbb{Z}} \def\fq{\lfloor q \rfloor} \def\fr{\lfloor r \rfloor} \def\fx{\lfloor x \rfloor} \def\fy{\lfloor y \rfloor} \def\fz{\lfloor z \rfloor} \Q_k$ denote $\Q\cap[k,k+1)$, and let $c:\Q_0^2\to\N$ be a bijective function counting off the elements of $\Q_0^2$.
Let superscripts from $\Z$ denote a simply transitive action of $\Z$ on $\Q_0$, and define $p:\Q^2\to\Q_0^2$ piecewise by $$p(x,y)=(\{x\}^{\fx-\fy},\{y\}^{\fx-\fy}).$$
Then the claim is satisfied with $s=c\circ p$.
Proof: Since $c$ is bijective, it is enough to show that $p$ is bijective on $\Q_m\times\Q_n$ for integer $m,n$, and bijective on horizontal and vertical lines. The bijectivity on $\Q_m\times\Q_n$ is immediate from the inverse of the group action, and the proofs for horizontal and vertical lines are the same. So it suffices to prove bijectivity on vertical lines.
Injectivity on Vertical Lines: When $p(x,y)=p(x,z)$, the first coordinates are $$\{x\}^{\fx-\fy}=\{x\}^{\fx-\fz}.$$ By the simple transitivity, $\fy=\fz$. So $(x,y)$ and $(x,z)$ are both in $\Q_{\fx}\times\Q_{\fy}$, on which $p$ is bijective, and $y=z$.
Surjectivity on Vertical Lines: Suppose $q \in \Q$ and $x,y\in\Q_0$. Let $k$ be an integer such that $\{q\}^k=x$. Let $r=\fq - k + y^{-k}$. Then $$\fq-\fr=\fq-(\fq-k)=k$$ and as desired $$p(q,r)=(\{q\}^k,\{r\}^k)=(x,(y^{-k})^k) = (x,y).$$
$\qquad\qquad$ Countable SUDOKU
I'll provide three parts:
A Sudoku plate is an infinite countable family $\ \mathbf A\ $ of infinite countable sets such that for arbitry finite $\ T\subseteq\bigcup\mathbf A,\ $ and $\ X\in\mathbf A,\ $ we have $$ X\setminus\bigcup\{Y\in\mathbf A\setminus\{X\}: Y\cap T\ne\emptyset\}\,\ \ne\ \ \emptyset $$
Sudoku solution definition: a solution is an arbitrary function $\ f:\bigcup\mathbf A\to\mathbb N\ :=\ \{1\ 2\ \ldots\} $ such that $\ f|X:X\to\mathbb N\ $ is a bijection for every $\ X\in\mathbf A.$
Solution: Define $$ \forall_{n\in\mathbb N}\qquad B_n:=\{n\in\mathbb N: n\equiv 2^{n-1}\mod 2^n\} $$
Let $\ \mathbf A=(X_n: n\in\mathbb N).$
Let $\ \pi:\bigcup\mathbf A\to\mathbb N\ $ be defined piecewise as a common extension of certain bijections
$$ \pi_n:\ X_n\setminus\bigcup_{m<n}X_m\ \to\ B_n $$ for every $\ n\in\mathbb N\ $ (thus, $\ \pi:\bigcup\mathbf A\to\mathbb N\ $ is a bijection).
Now, consider certain bijections:
$$ \rho_n:\ B_n\ \to\ \mathbb N\setminus \bigcup_{m<n}\left\{\rho_m(\pi(x)): x\in X_m\cap X_n\right\} $$ -- Solution -- $\ f:\bigcup\mathbf A\to\mathbb B\ $ can be given by:
$$ \forall_{n\in\mathbb N}\qquad \left( x\in X_n\setminus\bigcup\{X_m:m<n\}\quad\ \Rightarrow\quad f(x)\ =\ (\rho_n\circ\pi_n)(x) \right) $$
! Enjoy !
REMARK: The subtle moment is hidden in the selection of the ordering of $\ \bigcup\mathbf A$.
