Constructing a set that contains at most one point on vertical and horizontal I'm not sure how to answer this question:

Construct a set $A$, which is a subset of $[0, 1] \times [0, 1]$, such that $A$ contains at most one point on the horizontal and vertical lines, but boundary $(A) = [0, 1]\times [0, 1]$.

Is the answer just the open rectangle $(0, 1) \times (0, 1)$? How do you prove that the boundary of an open rectangle is a closed rectangle?
Thanks.
 A: The boundary of the open rectangle $(0,1)\times(0,1)$ is not $[0,1]\times[0,1]$. It is $\{0\}\times[0,1]\cup\{1\}\times[0,1]\cup[0,1]\times\{0\}\cup[0,1]\times\{1\}$. Also the open rectangle contains far more than one point on each horizontal and vertical line excluding the boundary lines.
To construct the desired set $A$ first pick a point with both coordinates being rational and in $[0,1]\times[0,1]]$. Subdivide $[0,1]\times[0,1]$ into four squares by forming the lines $\{\frac{1}{2}\}\times[0,1]$ and $[0,1]\times\{\frac{1}{2}\}$. Pick from each square an element such that both coordinates are rational and such that they don't lie a horizontal line or vertical line that can connect them to the previously chosen points. Continue subdividing into quarters as before and repeating the procedure. This process can be continued infinitely many times since there are infinitely many rationals and since at any finite stage we have only made finitely many choices and hence infinitely many lines are still available. The resulting set is dense by construction and has no interior since it is totally disconnected. The boundary will have the desired property.
