Question: Is it true that if two functions are not $ L^1$ functions, then their product is not a $ L^1$ function? How can we prove it?

Answer: 
I don't think so. Let $\theta(x)=\cases{1 \quad x>0\\0\quad x\leq0}$

we have $\int dx\, \theta(\pm x)=\infty$, but $\theta(x)\theta(-x)=0$ which is clearly $L^1$
It's false. Consider $f(x)=0$ if $x < 1$ and $f(x)=1/x$ if $x\geq 1$. Then $f \not \in L^1(\mathbb{R})$ but $f^2 \in L^1(\mathbb{R})$.