6. As shown in Figure 1, quadrilateral $ABCD$ is a unit square, $BD=BE$, $CE \parallel BD$. Then the length of $CE$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{3}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{6}-\sqrt{2}}{2}$
6. D.

As shown in Figure 3, draw $C O \perp$
$B D$ at point $O$, and $E F \perp B D$
at point $F$. Then $O$ is the center of the square $A B C D$.

It is easy to see that quadrilateral $C E F O$ is a rectangle.
$$
\text { Also, } B E=B D=\sqrt{2}, B O=C O=E F=\frac{\sqrt{2}}{2} \text {, }
$$

Thus, the three interior angles of Rt $\triangle B E F$ are $30^{\circ} 、 60^{\circ} 、 90^{\circ}$.
Therefore, $B F=\frac{\sqrt{3}}{2} B E=\frac{\sqrt{3}}{2} \times \sqrt{2}=\frac{\sqrt{6}}{2}$,
$$
C E=O F=B F-O B=\frac{\sqrt{6}-\sqrt{2}}{2} \text {. }
$$