There are several approaches to this proof. You could observe that any odd number of bit errors change the overall parity of the message which is detectable. Indeed you only need a single parity bit for the whole message to detect all errors consisting of an odd number of bits. You might be more explicit in examining the cases: if all 3 bits are in a single row, clearly the parity for that row is wrong. If all 3 bits are not in a single row, then at least one row has only 1 wrong bit and the parity for that row is wrong.