Heh, brute force works easily. Suppose that we have a $3 \times 3$ square with three corner squares of the same color, WLOG black. Then that square looks like this: B.B ... B.. We can now fill the rest of the square (forced by the no three colored squares in a row): BWB WWB BBW This square is symmetric with the top-left-to-bottom-right diagonal. As $n \ge 5$, this square must border at least two other sides (otherwise either the board is not a square or $n = 3 < 5$). Suppose it extends either upward or leftward (WLOG upward because of said symmetry): ..o BWB WWB BBW But then the upper-right square (marked o) cannot be colored properly (either a black line downward or a white line diagonally bottom-leftward). So this square extends to the right and bottom sides. Since $n \ge 5$, this square extends at least two units away, otherwise $n = 4 < 5$. BWB.. WWB2. BBW.. .1.B. ..... 1 and 2 cannot be both white (diagonally up-rightward white line), so at least one of them is black. WLOG 1 is black (due to symmetry). BWB.. WWB.. BBW.. .B.B. ..o.. But then the square marked o cannot be colored properly. The square above the o must be white (otherwise horizontal black line), and so the square with o will either make a diagonally up-leftward black line or vertical white line. So our assumption that there exists a $3 \times 3$ square with three corners of the same color is wrong. The conclusion follows.

haha this is such a cute problem! It basically turns into a game of connect-3