Consider a coordinate system, such that each square has coordinates $(x,y), x,y\in\mathbb{Z}$. WLOG, assume the square $(0,0)$ is black and $(0,0)\in P$ where by $P\subset \mathbb{Z}^2$ we denote the set of all squares inside the given polygon. For any $v\in\mathbb{Z}^2$ by $v+P$ we denote the translation of $P$ by vector $v$, i.e. the set $\{ v+p: p\in P\}$. Let $B$ be the set of all black squares in the plane and consider the family of polygons $\mathcal{P}:= \{b+P : b\in B \}$. Claim. The sets of the family $\mathcal{P}$ cover $\mathbb{Z}^2$ as required. Proof. First, we prove any two distinct sets $P_1,P_2\in \mathcal{P}$ are disjoint. Let $P_1=b_1+P,P_2=b_2+P$. Suppose on the contrary $\mathbb{Z}^2 \ni z_0\in P_1\cap P_2$. Consider the set $$X:= \{b_1+v : v\in \mathbb{Z}^2, z_0\in v+P_1 \}.$$ Note that $X$ is a translation of $P$, i.e. there exists a vector $v\in\mathbb{Z}^2$ with $X=v+P$ (a picture would be helpful). It easily follows, $b_1,b_2\in X$, a contradiction since both $b_1,b_2$ are black squares. Now, let's see there isn't uncovered square by the family $\mathcal{P}$. Seeking a contradiction, suppose it not true, i.e. there exists $z_0\in\mathbb{Z}^2$ such that $z_0\notin P', \forall P'\in\mathcal{P}$. Consider the seT $$Y:= \{ b_1+v : v\in\mathbb{Z}^2, z_0\in v+P_1\}.$$ Again, $Y$ is some translation of $P$. Then, there exists a black square $b\in Y$. But $z_0\in b+P$, a contradiction. $\blacksquare$