First we show we can pick 4 rows with at least 8 black squares. This is equivalent to finding 4 rows with 4 or less black squares. Suppose this were not possible, that is, every 4 rows combined had 5 or more black squares. Suppose we take the row with the least number of black squares, then the row with the second least number of black squares, and so on. Once we reach the fourth least row, there must have been two black squares on that row, otherwise the four rows combined cannot have 5 or more black squares. Then all the other eight rows must have two or more black squares as well, contradicting that there are only 12 black squares. Thus, we can find 4 rows with at least 8 black squares. Next, there are 4 or fewer black squares that are not chosen, so we may simply choose the 4 columns (or less) that contain those black squares.

Let us call the lines mentioned here as columns and define raws and columns together as lines. Just note that there are at least $4$ lines with at least $2$ black squares so choosing them we will have at most $12-4\dot 2=4$ remaining points which we can take out with the remaining $4$ lines. (If a black square lies in $2$ lines with at least $2$ points we just choose one of them).

Since there are 12 black squares and only 8 rows, according to Dirichlet's principle, we can locate at least 8 of the black squares in 4 rows. From there we are left with maximum of 4 black squares which we can locate with maximum of 4 lines. We showed that the 12 black squares can be located in 4 rows and 4 lines or less.

JBMO Sl 2018 easier version