6. As shown in Figure 1, the squares $A B C D$ and $A B E F$ with side length 2 are in planes that form a $60^{\circ}$ angle, and points $M$ and $N$ are on segments $A C$ and $B F$ respectively, with $A M=F N$. The range of the length of segment $M N$ is ( ).
(A) $\left[\frac{1}{2}, 2\right]$
(B) $[1,2]$
(C) $[\sqrt{2}, 2]$
(D) $[\sqrt{3}, 2]$
6. B.

As shown in Figure 2, draw $M H / / B C$ intersecting $A B$ at point $H$. Then
$$
\frac{A M}{A C}=\frac{A H}{A B} .
$$

Since $A M=F N$,
$$
A C=F B,
$$

thus, $\frac{F N}{F B}=\frac{A H}{A B} \Rightarrow N H / / A F$.
Therefore, $N H \perp A B, M H \perp A B$.
Hence, $\angle M H N=60^{\circ}$.
Let $A H=x(0 \leqslant x \leqslant 2)$. Then
$$
M H=x, N H=2-x \text {. }
$$

Thus, $M N=\sqrt{x^{2}+(2-x)^{2}-2 x(2-x) \cos 60^{\circ}}$
$$
=\sqrt{3(x-1)^{2}+1} \in[1,2] .
$$