Selection and Training Session

Let $O=AC\cap BD$; and, without loss of generality, $BO\ge DO$, $CO\ge AO$. If $AB\parallel DC$ then $1\le \frac{BO}{DO}=\frac{AO}{CO}\le 1$, whence $BO=OD$ and $ABCD$ is a parallelogram. Then by the problem condition $AC\le 1$, $BD\le 1$, hence $S(ABCD)\le \frac{1}{2}AC\cdot BD\le \frac{1}{2}<\frac{2}{3}$.

Let now $AB\nparallel DC$, and let $M$ be the intersection point of the rays $BA$ and $CD$; denote $BC=b$, $\angle ABC=\beta$, $\angle DCB=\gamma$. Then $S=S(ABCD)<S(BMC)$, while $$\begin{gathered} S(BMC)=\frac{1}{2}MC\cdot MB\sin (\beta +\gamma )=\frac{1}{2}\cdot \frac{b\sin \beta }{\sin (\beta +\gamma )}\cdot \frac{b\sin \gamma }{\sin (\beta +\gamma )}\cdot \sin (\beta +\gamma )= \\ =\frac{1}{2}\cdot \frac{b^2}{\mathrm{ctg}\beta +\mathrm{ctg}\gamma }, \end{gathered}$$ hence $$S\cdot (\mathrm{ctg}\beta +\mathrm{ctg}\gamma )<\frac{b^2}{2}.(1)$$ If $A_1\in [AB]$ and $D_1\in [CD]$ are the points such that $S(B{A}_{1}C)=\frac{1}{2}S=S(B{D}_{1}C)$ then by the problem condition $C{A}_1\le 1$, $B{D}_1\le 1$. Thus $$C{A}_1^2=b^2+\frac{S^2}{b^2{\sin }^2\beta }-2S\cdot \frac{\cos \beta }{\sin \beta }={\left(b-\frac{S}{b}\mathrm{ctg}\beta \right)}^2+\frac{S^2}{b^2}\le 1,$$ and, similarly, ${\left(b-\frac{S}{b}\mathrm{ctg}\gamma \right)}^2+\frac{S^2}{b^2}\le 1$. Then, in view of (1), $$\begin{gathered} 2\ge \frac{2S^2}{b^2}+\frac{1}{2}{\left(2b-\frac{S}{b}\left(\mathrm{ctg}\beta +\mathrm{ctg}\gamma \right)\right)}^2>\frac{2S^2}{b^2}+\frac{1}{2}{\left(2b-\frac{b}{2}\right)}^2= \\ =\frac{2S^2}{b^2}+\frac{9b^2}{8}\ge \frac{1}{2}\sqrt{\frac{2S^2}{b}\cdot \frac{9b^2}{8}}=3S. \end{gathered}$$ It follows that $S<\frac{2}{3}$.