### Proof for $(1)$:

In $\triangle CDB$ with $CD \perp AB$, we have:
- $\sin B = \frac{CD}{a}$.

Given that $\sin C = 2\sin A \sin B$, we can write:
- $\frac{\sin C}{\sin A} = 2\sin B$.

Substituting $\sin B$ from above, we get:
- $\frac{\sin C}{\sin A} = 2 \cdot \frac{CD}{a}$.

By the Law of Sines in $\triangle ABC$, we have $\frac{c}{\sin C} = \frac{a}{\sin A}$, which implies:
- $\frac{c}{a} = \frac{\sin C}{\sin A} = 2 \cdot \frac{CD}{a}$.

Solving for $CD$, we find:
- $CD = \frac{1}{2}c$.

Thus, we have proven that $CD = \frac{1}{2}c$, as required.
$\boxed{CD = \frac{1}{2}c}$

### Proof for $(2)$:

In $\triangle ABC$, the area $S_{\triangle ABC}$ can be expressed as $\frac{1}{2}ab\sin C$ and also as $\frac{1}{2}c \cdot CD$. From $(1)$, we know that $CD = \frac{1}{2}c$, which gives us:
- $c^{2} = 2ab\sin C$.

Using the Law of Cosines in $\triangle ABC$, we have:
- $c^{2} = a^{2} + b^{2} - 2ab\cos C$.

Given that $a^{2} + b^{2} = \sqrt{6}ab$, we can write:
- $2ab\sin C = \sqrt{6}ab - 2ab\cos C$,
- $\sin C + \cos C = \frac{\sqrt{6}}{2}$.

Using the identity $\sqrt{2}\sin\left(C + \frac{\pi}{4}\right) = \sin C + \cos C$, we get:
- $\sqrt{2}\sin\left(C + \frac{\pi}{4}\right) = \frac{\sqrt{6}}{2}$,
- $\sin\left(C + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}$.

Considering the range of $C + \frac{\pi}{4}$, we have two possible solutions for $\sin\left(C + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}$, which are:
- $C + \frac{\pi}{4} = \frac{\pi}{3}$ or $C + \frac{\pi}{4} = \frac{2\pi}{3}$.

Solving for $C$, we find:
- $C = \frac{\pi}{12}$ or $C = \frac{5\pi}{12}$.

Given the geometric constraint that either $\angle ACD$ or $\angle BCD$ must be greater than or equal to $\frac{\pi}{4}$, we conclude:
- $C = \frac{5\pi}{12}$.

Therefore, the angle $\angle ACB$ is $\boxed{\frac{5\pi}{12}}$.