6. As shown in Figure 1, the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. On the side diagonal $A_{1} D$, take point $M$, and on $C D_{1}$, take point $N$, such that line segment $M N$ is parallel to the diagonal face $A_{1} A C C_{1}$. Then the minimum length of such $M N$ is $(\quad)$.
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$
6. A.

As shown in Figure 4, draw $M M_{1} \perp A D$ at point $M_{1}$, and $N N_{1} \perp D C$ at point $N_{1}$.
It is easy to prove that
$$
M_{1} N_{1} \parallel A C \text{. }
$$

Let $D M_{1}$
$$
=D N_{1}=x \text{. }
$$

Then $M M_{1}=x$,
$$
N N_{1}=1-x \text{. }
$$

Draw $M H \perp N N_{1}$ at point $H$. Then
$$
N H=1-2 x, M_{1} N_{1}=\sqrt{2} x \text{. }
$$

By the Pythagorean theorem, we get
$$
M N^{2}=(\sqrt{2} x)^{2}+(1-2 x)^{2}=6\left(x-\frac{1}{3}\right)^{2}+\frac{1}{3}
$$

When $x=\frac{1}{3}$, $M N$ reaches its minimum value $\frac{\sqrt{3}}{3}$.