Weak continuous Sudoku:
A weak continuous Sudoku can be constructed based on the ideas that you already provided.
First, we construct a weak continuous Sudoku for the set $U=(0,1]$ instead of $U=[0,1]$.
Here, a weak continuous Sudoku can be constructed by using the function
$f$ from your attempt but as a function $f:(0,1]^2\to (0,1]$ (since one boundary is gone, the problems that you observed are now gone, too).
Then, choose a bijection $h:[0,1]\to (0,1]$
(an explicit bijection can be constructed if you prefer a constructive soution).
Then we define
$$
g:[0,1]^2\to [0,1],
\qquad
(x,y)\mapsto h^{-1} (f(h(x),h(y))).
$$
This function $g$ then can be shown to be a weak continuous Sudoku.
Strong continuous Sudoku:
As for strong continuous Sudoku, things get more complicated
and it would be a lot of work to explain my construction in full detail,
but I can provide a sketch.
First, the bijection $h$ above should be chosen such that
each interval in $[0,1]$ contains a subinterval $[ a,b ]$ such that $h(x)=x$
for all $x\in[a,b]$, see the comments below for such a construction.
Furthermore, it uses a bijection $j:[0,1]\to [0,1]$
such that $j((c,d))$ is dense in $[0,1]$ for all intervals $(c,d)$,
see the comments below for such a construction for $j$.
Then one can mix the rows or columns of the previous weak Sudoku according to $j$,
i.e. $\tilde g(x,y)=g(j(x),y)$.
This function $\tilde g$ should then be a strong continuous Sudoku.
Let me provide a rough sketch how this can be done.
Let $S$ be a square sub-region of $[0,1]^2$.
Let $S_2=[a,b]\times [c,d]\subset S$ be a smaller square sub-region,
where $a<b,c<d$ are such that
$h(x)=x$ holds for all $x\in[a,b]\cup[c,d]$
(such a sub-region exists due to the comments above on the choice of $h$).
It suffices to show that $\tilde g(S_2)=[0,1]$ instead of $\tilde g(S)=[0,1]$.
Let $t\in [0,1]$ be given.
Let $m:=(c+d)/2$.
Since $j([a,b])$ is dense in $[0,1]$,
the function values $\{\tilde g(x,m)| x\in[a,b]\}$ are also dense in $[0,1]$.
Let $s\in[a,b]$ be such that $\tilde g(s,m)$ is close to $t$ in the sense that
$$
t-\frac{d-c}{2} < \tilde g(s,m) < t+\frac{d-c}{2}.
$$
By exploiting the definitions of $\tilde g,g,f$ we have
$\tilde g(s,m+x)=\tilde g(s,m)+x$
for $x\in (-\frac{d-c}{2},\frac{d-c}{2})$
(with the exception that the values wrap around at $1$).
By setting $x=\tilde g(s,m)-t$,
we get $t=\tilde g(s,m+x)$ and $(s,m+x)\in S_2 = [a,b]\times [c,d]$.
Thus $t$ can be reached and the condition (5.) for strong continuous Sudoku
is satisfied.
on the existence of a function $h$:
We can define $h:[0,1]\to (0,1]$ by setting $h(0)=1/2$, $h(1/2)=1/3$, $h(1/3)=1/4$,
etc., and $h(x)=x$ for all other $x$.
Then for each interval one can find a sufficiently small subinterval
$[a,b]$ such that $h(x)=x$ for all $x\in[a,b]$.
on the existence of a function $j$:
This is more complicated, so let me provide a rough sketch.
Let $(q_k)_k$ be an enumeration of the rational numbers in $[0,1]$
and let $I_k$ be an interval of length $2^{3-2k}$ centered at $q_k$.
We define the sets
$$ A_k := I_k\setminus \bigcup_{l>k} I_l.$$
These sets form a partition of $[0,1]$ and each set $A_k$ has cardinality
equal to $[0,1]$.
Let $(B_k)_k$ be another sequence of subsets of $[0,1]$ which
form a partition of $[0,1]$ such that each $B_k$ is dense and has
cardinality equal to $[0,1]$
(such a partition exists, one can append dense countable sets with enough other elements to form sets $B_k$,
but I think this requires the axiom of choice).
Then we construct $j$ by (bijectively) mapping
$A_k$ to $B_k$.
Since the lengths of the sets $A_k$ get smaller and smaller
and the rationals $q_k$ are dense,
each interval has a subinterval of the form $I_k$.
Since $I_k$ contains $A_k$ and $A_k$ is mapped to a dense set $B_k$,
we obtain the desired property that $j(I_k)$ is dense in $[0,1]$.
Best Answer
Remember that first-order logic has no idea what a 'digit' is, so you'll need to specify that:
$\forall x (x=1 \lor x=2 \lor x=3 \lor x=4 \lor x=5 \lor x=6 \lor x=7 \lor x=8 \lor x = 9)$
In fact, even that is not enough, since just because you have different symbols does not mean that they should denote different objects. So, you'll actually have to specify that these are all different, i.e. you need:
$1 \not = 2$, $1 \not = 3$, etc.
Even then, first-order has no idea that $1$ is 'next' to $2$, etc. ... but you'll need that knowledge to indicate the blocks .. or at least that $1$,$2$, and $3$ go together, and same for $4$, $5$, $6$, and for $7$, $8$, $9$
So, maybe have a two-place $b$ relation to indicate that:
$b_{1,2}$, $b_{2,3}$, $b_{4,5}$, $b_{5,6}$, $b_{7,8}$, $b_{8,9}$
and take its symmetric and transitive closure:
$\forall z_1 \forall z_2 (b_{z_1,z_2} \to b_{z_2,z_1})$
$\forall z_1 \forall z_2 \forall z_3 ((b_{z_1,z_2} \land b_{z_2,z_3})\to b_{z_1,z_3})$
And now you can specify that there cannot be two of the same digits in the same block:
$\forall x_1 \forall x_2 \forall y_1 \forall y_2 \forall v ((a_{x_1,y_1,v} \land \neg (x_1 = x_2 \land y_1 = y_2) \land b_{x_1,x_2} \land b_{y_1,y_2}) \to \neg a_{x_2,y_2,v})$
Oh, and finally, make sure to say that every square needs a digit!
$\forall x \forall y \exists v \ a_{x,y,v}$