7. Given the curve $C: y=\sqrt{-x^{2}-2 x}$ and the line $l: x+y-m=0$ have two intersection points, then the range of $m$ is ( ).
A. $(-\sqrt{2}-1, \sqrt{2})$
B. $(-2, \sqrt{2}-1)$
C. $[0, \sqrt{2}-1)$
D. $(0, \sqrt{2}-1)$
7. C. The curve $C$ is given by $(x+1)^{2}+y^{2}=1(y \geqslant 0)$, which represents the upper half (including endpoints) of a circle centered at $(-1,0)$ with a radius of 1. $l$ is a line with a slope of -1 and an intercept of $m$.
From the fact that the line $l$ intersects the curve $C$ at two points, we know that
$$
\frac{|-1-m|}{\sqrt{2}}<1 \text {, and } m \geqslant 0 \text {. }
$$

Solving this, we get $0 \leqslant m<\sqrt{2}-1$.