Maximum possible variance

It is true. A simple reason is that $$\mathrm{var}(X)=E(X^2)-m^2$$ where $m=E(X)$ and that, if $X$ is almost surely in $[0,1]$, then $X^2\leqslant X$ almost surely hence $E(X^2)\leqslant E(X)=m$, thus $$\mathrm{var}(X)\leqslant m-m^2=m(1-m). $$ More generally, if $0\leqslant X\leqslant x$ almost surely then $\mathrm{var}(X)\leqslant xm-m^2$.