A. False.

For $A$ to be open, for every point $p \in A$ there must be an open set $Up$ such that $p \in Up \subset A$.

Consider the point $(0,0) \in A$ and let $U_p$ be and open set containing $(0,0)$. We will show that $U_p$ is not fully contained in $A$.

Since $U_p$ is open, we can build an open ball $B_{r}(0,0)$ of radius $r$ centered at $(0,0)$ that is fully contained in $U_p$. If we take $n$ large enough, for example $n>\max\{\frac{1}{r},1\}$, the point $(\frac{1}{n},0) \in B_{r}(0,0) \subset U_p$. But we know that $(\frac{1}{n},0) \notin A$, since $n >1$. This means that $U_p \not \subset A$. Since $U_p$ was arbitrary, no open set containing $(0,0)$ can be fully contained in $A$.

B. True.

We can show that $A^c$ is open. Let $(p,q) \in A^c$ and let $(a,b)$ be the closest point of $A$ to $(p,q)$. Such closest point exists since either $a= \lfloor p \rfloor$ or $a = \lceil p\rceil$, $b= \lfloor q \rfloor$ or $b = \lceil q \rceil$. Any other point in $A$ would have a larger distance to $(p,q)$.

Let $r =d\left((p,q),(a,b)\right)>0$, since either $p \neq a$ or $q \neq b$. The ball centered around $(p,q)$ of radius $r$ is fully contained in $A^c$:

Take any $(s,t) \in B_r(p,q)$. We will show that $d((s,t),(n,m))>0$ for any point $(n,m) \in A$, i.e, $(s,t)$ does not lie in $A$.

Suppose $d((s,t),(n,m))=0$, then $d\left((p,q),(n,m)\right) \leq d((p,q),(s,t)) + d((s,t),(n,m)) < r+ 0 $, a contradiction, since $r$ is the distance from $(p,q)$ to the nearest point in $A$.