Let $x_1, x_2, \ldots, x_8$ be the number of black squares in the rows, such that $x_1 \ge x_2 \ge \ldots \ge x_8$. This means that if, for example, the fourth line has the maximum number of black squares, then these are $x_1$. From the problem condition we have:

$$x_1 + x_2 + \ldots + x_8 = 12. \quad (1)$$

We observe that it is impossible $x_1 = x_2 = \ldots = x_8$, since then $8x_1 = 12$, a contradiction. This means that:

$$x_1 + x_2 + x_3 + x_4 > x_5 + x_6 + x_7 + x_8.$$

We will prove that it is impossible to have:
$$x_1 + x_2 + x_3 + x_4 = x_5 + x_6 + x_7 + x_8 + 1,$$
Since then (1) gives $2(x_1 + x_2 + x_3 + x_4) - 1 = 12$, a contradiction, since the left hand side is odd, while the right hand side is even. Moreover, we cannot have:
$$x_1 + x_2 + x_3 + x_4 = x_5 + x_6 + x_7 + x_8 + 2. \qquad (2)$$
Indeed, if (2) was true, then from (1) we will have that $x_1 + x_2 + x_3 + x_4 = 7$ and
$$x_5 + x_6 + x_7 + x_8 = 5, \qquad (3)$$
so that $4x_4 \le 7 \Rightarrow x_4 \le 1$. From the ordering condition we have $x_8 \le x_7 \le x_6 \le x_5 \le x_4 \le 1$, so $x_8 + x_7 + x_6 + x_5 \le 4$, which is a contradiction according to (3).
Therefore:
$$x_1 + x_2 + x_3 + x_4 \ge x_5 + x_6 + x_7 + x_8 + 3,$$
so from (1) we have:
$$2(x_1 + x_2 + x_3 + x_4) - 3 \ge 12 \Rightarrow x_1 + x_2 + x_3 + x_4 \ge 8,$$
thus, if we choose the $4$ lines with the maximum number of black squares, then these contain at least $8$ black squares. Then, we are left with at most $4$ black squares and for them we choose the $4$ columns in which they are located. As a conclusion, with $4$ lines and $4$ columns we can cover $12$ black squares.