6. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle A=$ $60^{\circ}, A C=1, D, E$ are on sides $B C, A B$ respectively, such that $\triangle A D E$ is an isosceles right triangle, $\angle A D E=90^{\circ}$. Then the length of $B E$ is ( ).
(A) $4-2 \sqrt{3}$
(B) $2-\sqrt{3}$
(C) $\frac{\sqrt{3}-1}{2}$
(D) $\sqrt{3}-1$
6. A.

As shown in Figure 3, draw $E F \perp B C$ at point $F$. It is easy to see that
$\triangle A C D \cong \triangle D F E$,
$\triangle E F B \backsim \triangle A C B$.
Let $E F=x$. Then
$$
\begin{array}{l}
B E=2 x, A E=2-2 x, \\
D E=\sqrt{2}(1-x), D F=A C=1 .
\end{array}
$$

Thus, $1^{2}+x^{2}=[\sqrt{2}(1-x)]^{2}$
$$
\Rightarrow x^{2}-4 x+1=0 \text {. }
$$

Given $0<x<1$, we get $x=2-\sqrt{3}$.
Therefore, $B E=2 x=4-2 \sqrt{3}$.