9. $[1, \sqrt{2}]$.

From the condition, we get $0 \leqslant x \leqslant 1$. Let $\sin \theta=\sqrt{x}, \theta \in\left[0, \frac{\pi}{2}\right], \sqrt{1-x}=\cos \theta$, then $y=\sin \theta+ \cos \theta=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)$. Therefore, $1 \leqslant y \leqslant \sqrt{2}$.