Consider this system of equations $$ \begin{cases} xy=a\\ x+y=a\\ 0\leq x\leq 1\\ 0\leq y\leq 1\\ \end{cases} $$ where $a\in [0,1]$ is a parameter and $x,y$ are the unknowns.
When $a=0$, $x=0,y=0$ are clearly the solutions of the system. I think that there are no solutions for other values of $a$. Is it enough to show that by rewriting the system as $$ y^2-ay+a=0 $$ and then saying that, for any $a\in (0,1]$, $a^2-4a<0$?
$$(x-y)^2=(x+y)^2-4xy=a(a-4)\ge0$$ is indeed contradictory with
$$0\le xy=a\le 1$$ except for $a=0$.