Proof:

Part (1): Given $c-2b\cos A=b$, we start by applying the Law of Sines to translate this into an equation involving sines:

$c - 2b\cos A = b$

Applying the Law of Sines and substituting for $c$ and $b$ in terms of $\sin C$ and $\sin B$ respectively, we get:

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

Using the identity for $\sin(A + B) = \sin A\cos B + \cos A\sin B$, and knowing $A + B + C = \pi$ (sum of angles in a triangle), we have $\sin C = \sin (A + B)$. Plugging this into our equation:

$\sin (A + B) - 2\sin B\cos A = \sin B$

$\Rightarrow \sin A\cos B + \cos A\sin B - 2\sin B\cos A = \sin B$

Rearranging and simplifying gives us:

$\sin(A - B) = \sin B$

Given $A, B \in (0, \frac{\pi}{2})$ (since $\triangle ABC$ is acute), we know $A - B \in (-\frac{\pi}{2}, \frac{\pi}{2})$. In this interval, $\sin x$ is monotonically increasing, therefore:

$A - B = B \Rightarrow A = 2B$

Hence, we have proved that $A = 2B$. $\boxed{A = 2B}$

Part (2): With $A = 2B$ established, we now consider the angle bisector of $A$ intersecting $BC$ at $D$. Given $\angle ABC = \angle BAD$ and using the Law of Sines in $\triangle ABD$, we can express $AD$ in terms of $B$:

$\frac{AD}{\sin B} = \frac{AB}{\sin(\pi - 2B)} = \frac{2}{\sin 2B}$

This gives us $AD = BD = \frac{1}{\cos B}$ (since $AB = c = 2$ and using the double angle formula for sine).

The area of $\triangle ABD$ is then calculated as:

$S_{\triangle ABD} = \frac{1}{2}AB \times AD \times \sin B = \frac{1}{2} \times 2 \times \frac{1}{\cos B} \times \sin B = \tan B$

Given the acute nature of $\triangle ABC$, we establish bounds for $B$:

$0 < B < \frac{\pi}{2}$, $0 < 2B < \frac{\pi}{2}$, and $0 < \pi - 3B < \frac{\pi}{2}$

Solving these inequalities gives us $\frac{\pi}{6} < B < \frac{\pi}{4}$. Hence, the range of $\tan B$ is $\left(\tan \frac{\pi}{6}, \tan \frac{\pi}{4}\right) = \left(\frac{\sqrt{3}}{3}, 1\right)$.

Therefore, the range of values for the area of $\triangle ABD$ is $\boxed{\left(\frac{\sqrt{3}}{3}, 1\right)}$.