Among the four pairs of numbers $(x, y)$ formed from the values 0 and 1, there is at least one for which (1) holds. Indeed, for these pairs, the value of the function $f(x)+g(y)-xy$ is

$$\begin{aligned} a & =f(0)+g(0) \\ b & =f(0)+g(1) \\ c & =f(1)+g(0) \\ d & =f(1)+g(1)-1 \end{aligned}$$

Thus,

$$b+c-a-d=1$$

and therefore, among the numbers $b, c, (-a), (-d)$, at least one has an absolute value of at least 1/4. This completes the proof of our statement.

Remark. If in the hypothesis, the closed interval $[0,1]$ for $x$ and $y$ is replaced by the open interval $(0,1)$, then the statement of the problem is no longer true. An example of this is

$$f(x)=g(x)=\frac{x}{2}-\frac{1}{8}$$