THE PROBLEM :
Source : Rohatgi, Saleh. p.$114$. Problem $16$.
This is my attempt to construct a solution. I'd be grateful if anyone checks whether or not it is technically alright. Also, I'd like to request to provide any shorter method, if available, to solve the problem. Thanks in advance.
I use the shorthand $\int_{a}^{b} \int_{c}^{d}$ to mean $\int_{a}^{b} \int_{c}^{d} f(x,y)dxdy$
Then, $F_1(a)F_2(b)=\int_{-\infty}^{\infty} \int_{-\infty}^{a} \times \int_{-\infty}^{b} \int_{-\infty}^{\infty} = (\int_{-\infty}^{b} \int_{-\infty}^{a}+\int_{b}^{\infty} \int_{-\infty}^{a}) \times (\int_{-\infty}^{b} \int_{-\infty}^{a}+\int_{-\infty}^{b} \int_{a}^{\infty})$
This breaks up into four components :
$(\int_{-\infty}^{b} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{-\infty}^{a})+(\int_{-\infty}^{b} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{a}^{\infty})+(\int_{b}^{\infty} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{-\infty}^{a})+(\int_{b}^{\infty} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{a}^{\infty})$
Adding and subtracting $(\int_{-\infty}^{b} \int_{-\infty}^{a})(\int_{a}^{\infty} \int_{b}^{\infty})$ and noting that probability measure sums up to $1$, we have
$(\int_{-\infty}^{b} \int_{-\infty}^{a})+(\int_{b}^{\infty} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{a}^{\infty})-(\int_{-\infty}^{b} \int_{-\infty}^{a})(\int_{b}^{\infty} \int_{a}^{\infty})$
The extra term is :
$(\int_{b}^{\infty} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{a}^{\infty})-(\int_{-\infty}^{b} \int_{-\infty}^{a})(\int_{b}^{\infty} \int_{a}^{\infty})$
We shall show that this is $\geq 0$ $[$which will give us the proof since $F(a,b)=(\int_{-\infty}^{b} \int_{-\infty}^{a})]$
Now, $(\int_{b}^{\infty} \int_{-\infty}^{a})(\int_{-\infty}^{b} \int_{a}^{\infty})=(\int_{b}^{\infty} \int_{-\infty}^{a}f(x_1,y_2)dx_1dy_2)(\int_{-\infty}^{b} \int_{a}^{\infty}f(x_2,y_1)dx_2dy_1)$
$=\int_{b}^{\infty} \int_{-\infty}^{a} \int_{-\infty}^{b} \int_{a}^{\infty} f(x_1,y_2) f(x_2,y_1) dx_2dy_1dx_1dy_2$
By the condition given in the question, this quantity is greater than
$\int_{b}^{\infty} \int_{-\infty}^{a} \int_{-\infty}^{b} \int_{a}^{\infty} f(x_1,y_1) f(x_2,y_2) dx_2dy_1dx_1dy_2$
$=\int_{-\infty}^{b} \int_{-\infty}^{a} \int_{b}^{\infty} \int_{a}^{\infty} f(x_1,y_1) f(x_2,y_2) dx_2dy_2dx_1dy_1$
$=(\int_{-\infty}^{b} \int_{-\infty}^{a} f(x_1,y_1) dx_1dy_1) \times (\int_{b}^{\infty} \int_{a}^{\infty} f(x_2,y_2) dx_2dy_2)=(\int_{-\infty}^{b} \int_{-\infty}^{a})(\int_{b}^{\infty} \int_{a}^{\infty})$
Hence the extra term is positive. Hence the proof.