Define directly that $\widetilde{f}(x)=lim f(a_{n})$ for $a_{n}\in X$ that $a_{n}\rightarrow x$.
For $(a_{n}),(b_{n})$ both $a_{n}\rightarrow x$ and $b_{n}\rightarrow x$, consider $c_{2n+1}=a_{n}$ and $c_{2n}=b_{n}$, then $c_{n}\rightarrow x$ and $d(f(c_{n}),f(c_{m}))\leq Cd(c_{n},c_{m})\rightarrow 0$. As $Y$ is complete, $f(c_{n})\rightarrow y$. But $(f(a_{n}))$ and $(f(b_{n}))$ are subsequences of $(f(c_{n}))$, so $y=\lim f(a_{n})=\lim f(b_{n})$ and hence it is well-defined.
Consider the constant sequence $(x,x,...)$, $x\in X$, then $\widetilde{f}(x)=f(x)$.
To prove the continuity, note that $d(\widetilde{f}(x),\widetilde{f}(y))=\lim d(f(a_{n}),f(b_{n}))\leq\lim d(a_{n},b_{n})=d(x,y)$ for $a_{n}\rightarrow x, b_{n}\rightarrow y$.
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
If you do a precalculus analysis of the graph of the function $g(x)=x^3-3x$ you'll see that $$g(-\sqrt{3},+\sqrt{3}) = [-2,+2] $$ Now just linearly scale and shift the $x$ and $y$ axes to adjust the domain and range: $$f(x) = a + b g(cx + d) $$ Letting $c = 2 \sqrt{3}$ and $d=-\sqrt{3}$, the function $cx+d$ takes $(0,1)$ to $(-\sqrt{3},\sqrt{3})$.
Then $g$ takes $(-\sqrt{3},+\sqrt{3})$ to $[-2,+2]$
And finally, letting $a=\frac{1}{2}$ and $b=\frac{1}{4}$, the function $a+by$ takes the interval $[-2,2]$ to the interval $[0,1]$.