### Solution:

Part (1) Proof:

Given the condition
$$\frac{\sin A}{\sin C} - 1 = \frac{\sin^2 A - \sin^2 C}{\sin^2 B}$$

Simplifying the left-hand side,
$$\frac{\sin A - \sin C}{\sin C} = \frac{\sin^2 A - \sin^2 C}{\sin^2 B}$$

We then get
$$\frac{1}{\sin C} = \frac{\sin A + \sin C}{\sin^2 B}$$

Applying the Law of Sines in $\triangle ABC$, we have
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

This leads to $b^2 = c^2 + ac$.

By the Law of Cosines,
$$b^2 = a^2 + c^2 - 2ac\cos B$$

Substituting $b^2$ from earlier,
$$c^2 + ac = a^2 + c^2 - 2ac\cos B$$

Simplifying gives
$$c = a - 2c \cos B$$

This implies
$$\sin C = \sin A - 2\sin C \cos B$$

Further simplification gives
$$\sin C = \sin (B + C) - 2\sin C \cos B$$

Which simplifies to
$$\sin C = \sin (B - C)$$

In an acute triangle, this leads to
$$C = B - C$$

Thus, we have
$$B = 2C$$
$\boxed{B = 2C}$

**Part (2) Finding the range of lengths of segment $BD$:**

In $\triangle BCD$, applying the Law of Sines,
$$\frac{4}{\sin \angle BDC} = \frac{BD}{\sin C}$$

Thus,
$$BD = \frac{4\sin C}{\sin \angle BDC} = \frac{4\sin C}{\sin 2C} = \frac{2}{\cos C}$$

Since $B = 2C$ and $\triangle ABC$ is an acute triangle,
$$0 < C < \frac{\pi}{2}, \quad 0 < 2C < \pi, \quad 0 < \pi - 3C < \frac{\pi}{2}$$

Solving gives
$$\frac{\pi}{6} < C < \frac{\pi}{4}$$

Hence,
$$\frac{\sqrt{2}}{2} < \sin C < \frac{\sqrt{3}}{2}$$

Therefore,
$$\frac{4\sqrt{3}}{3} < BD < 2\sqrt{2}$$

The range of lengths of segment $BD$ is
$$\boxed{\left(\frac{4\sqrt{3}}{3}, 2\sqrt{2}\right)}$$