How can one build a subset $A\subset [0,1]\times[0,1]$ containing at the most one point from each horizontal and each vertical section and whose boundary (frontier) is $[0,1]\times[0,1]$?
I don't know how to build this set. I know that if $A$ contains points in each quarter of the square it's enough.
Let $\alpha, \beta$ be two irrationals such that $\alpha/\beta$ is also irrational. What can one say about the set $$A=\{(n\alpha \mod 1,n\beta\mod 1):n\in \Bbb Z\}?$$
Edit: probably I should give some more hints:
Edit 2: As Thomas pointed out in the comment, Statement 2. above is not correct unless $n\in \Bbb Q$. One then can modify the set to $$A'=\{(r\alpha\mod 1,r\beta\mod 1):\color{red}{r\in\Bbb Q}\}.$$ Note that the original $A$ still works, just not with the given argument.
Expanding upon @QuangHoang's hint.
To show that $A'$ is dense in the unit square, it suffices to show that the image of the map $f:\mathbb{R} \to \mathbb{R}^2$ defined by $f(t) = (\alpha t \pmod 1,\beta t \pmod 1)$ is dense in $[0,1] \times [0,1]$. This is because $\mathbb{Q}$ is dense in $\mathbb{R}$, so $f(\mathbb{Q})$ is dense in $f(\mathbb{R})$, which is dense in $[0,1] \times [0,1]$ as per our claim. This implies that $f(\mathbb{Q})$ is dense in $[0,1] \times [0,1]$ and $f(\mathbb{Q}) = A'$ so we are done.
The proof that $f(\mathbb{R})$ is dense in $[0,1] \times [0,1]$ can be seen here.
Showing that $A$ is dense in $[0,1] \times [0,1]$ takes a bit more work. A proof can be found here.
You can also do this by constructing a function $f: [0, 1] \rightarrow [0, 1]$ such that $f$ is identity on irrationals, $f$ restricted to rationals is a bijection and the graph of $f$ is dense in the unit square. Just list a countable basis $\{B_n : n \geq 1\}$ and indutively ensure that the graph of $f$ picks a point from each $B_n$.
