jmc

First note that since points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$, we have $HG=GF=FE=AB=BC=CD=1$. Also, $\overline{HG}$ is parallel to $\overline{CD}$ and $HG=CD$, so $CDGH$ is a parallelogram. Similarly, $\overline{AB}$ is parallel to $\overline{FE}$ and $AB=FE$, so $ABEF$ is a parallelogram. As a consequence, $WXYZ$ is a parallelogram, and since $HG=CD=AB=FE$, it is a rhombus.



Since $AH=AC=2$, the rectangle $ACFH$ is a square of side length 2. Its diagonals $\overline{AF}$ and $\overline{CH}$ have length $2\sqrt{2}$ and form a right angle at $X$. As a consequence, $WXYZ$ is a square. In isosceles $△HXF$ we have $HX=XF=\sqrt{2}$. In addition, $HG=\frac{1}{2}HF$. So $XW=\frac{1}{2}XF=\frac{1}{2}\sqrt{2}$, and the square $WXYZ$ has area $X{W}^2=\boxed{\frac{1}{2}}$.