jmc

We claim that the minimum is $\frac{4}{9}.$ When $x=y=\frac{1}{3},$ $$\begin{array}{rl}xy& =\frac{1}{9},\\ (1-x)(1-y)& =\frac{4}{9},\\ x+y-2xy& =\frac{4}{9}.\end{array}$$The rest is showing that one of $xy,$ $(1-x)(1-y),$ $x+y-2xy$ is always at least $\frac{4}{9}.$

Note that $$xy+(1-x-y+xy)+(x+y-2xy)=1.$$This means if any of these three expressions is at most $\frac{1}{9},$ then the other two add up to at least $\frac{8}{9},$ so one of them must be at least $\frac{4}{9}.$

Let $s=x+y$ and $p=xy.$ Then $$s^2-4p=(x+y{)}^2-4xy=(x-y{)}^2\ge 0.$$Assume $x+y-2xy=s-2p<\frac{4}{9}.$ Then $$0\le s^2-4p<{\left(2p+\frac{4}{9}\right)}^2-4p.$$This simplifies to $81p^2-45p+4>0,$ which factors as $(9p-1)(9p-4)>0.$ This means either $p<\frac{1}{9}$ or $p>\frac{4}{9}$; either way, we are done.

Therefore, the maximum value is $\boxed{\frac{4}{9}}.$