6. Let points $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $l$ be the right directrix. If there exists a point $M$ on the ellipse such that $\left|M F_{1}\right|$, $\left|M F_{2}\right|$, and the distance $d$ from point $M$ to $l$ form a geometric progression, then the range of the eccentricity $e$ of the ellipse is ( ).
(A) $[\sqrt{2}-1,1)$
(B) $(0, \sqrt{2}-1]$
(C) $\left[\frac{\sqrt{2}}{2}, 1\right]$
(D) $\left(0, \frac{\sqrt{2}}{2}\right]$
6. (A).

From $\left|M F_{2}\right|^{2}=\left|M F_{1}\right| \cdot d$ we get $\frac{\left|M F_{1}\right|}{\left|M F_{2}\right|}=\frac{\left|M F_{2}\right|}{d}=e$, i.e., $\left|M F_{1}\right|=e\left|M F_{2}\right|$.
Also, $\left|M F_{1}\right|+\left|M F_{2}\right|=2 a$.
Solving (1) and (2) we get
$\left|M F_{1}\right|=\frac{2 a e}{1+e},\left|M F_{2}\right|=\frac{2 a}{1+e}$.
In the ellipse, we have $\left|M F_{2}\right|-\left|M F_{1}\right| \leqslant 2 c$,

i.e., $\frac{1-e}{1+e} \leqslant \frac{c}{a}=e$.
Thus, $\sqrt{2}-1 \leqslant e<1$.