Continuous Function

Solution 1:

Let $\varepsilon > 0$. There is $\delta_1 > 0$ such that $$f(x_0, y_0) - \varepsilon/2 < f(x_0 - \delta_1, y_0) < f(x_0 + \delta_1, y_0) < f(x_0, y_0) + \varepsilon/2.$$ Then there is $\delta_2 > 0$ such that $$f(x_0 - \delta_1, y_0) - \varepsilon/2 < f(x_0 - \delta_1, y) < f(x_0 - \delta_1, y_0) + \varepsilon/2$$ and $$f(x_0 + \delta_1, y_0) - \varepsilon/2 < f(x_0 + \delta_1, y) < f(x_0 + \delta_1, y_0) + \varepsilon/2$$ for any $y \in (y_0 - \delta_2, y_0 + \delta_2)$. Now for any $(x,y) \in (x_0 - \delta_1, x + \delta_1) \times (y_0 - \delta_2, y_0 + \delta_2)$ you have $|f(x,y) - f(x_0, y_0)| < \varepsilon$.