### Solution

#### 1. Proof

We start from the given equation:
$$\sin(B+C) = \dfrac{2S}{a^{2}-c^{2}}$$
Since $B+C=\pi-A$, we have:
$$\sin A = \dfrac{2S}{a^{2}-c^{2}}$$
By the Law of Sines, we know that $2S=bc\sin A$, hence:
$$\sin A = \dfrac{bc\sin A}{a^{2}-c^{2}}$$
Since $\sin A \neq 0$, we can cancel $\sin A$ on both sides, getting:
$$a^{2}-c^{2}=bc$$
Using the Law of Cosines for side $a$, we find:
$$a^{2}=b^{2}+c^{2}-2bc\cos A$$
Subtracting $c^2$ from both sides, we have:
$$a^{2}-c^{2}=b^{2}-2bc\cos A$$
Equating the two expressions for $a^{2}-c^{2}$:
$$b^{2}-2bc\cos A=bc$$
Rearranging gives:
$$b-2c\cos A=c$$
Substituting $\sin B$ for $b$ and $\sin C$ for $c$, we have:
$$\sin B-2\sin C\cos A=\sin C$$
Applying the sum-to-product identities:
$$\sin(A+C)-2\sin C\cos A=\sin C$$
Simplifying using trigonometric identities, we get:
$$\sin A\cos C-\cos A\sin C=\sin C$$
This simplifies to:
$$\sin(A-C)=\sin C$$
Given that $A$, $B$, $C \in (0,\pi)$, the only way for this to hold true is if $A=2C$. Thus, we have proven that $A=2C$.

#### 2. Range for $S$

Given $A=2C$, then $B=\pi -3C$, and thus:
$$\sin B=\sin 3C$$
Using the Law of Sines:
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B} \implies a=\dfrac{2\sin 2C}{\sin 3C}$$
The area $S$ can be expressed as:
$$S=\dfrac{1}{2}ab\sin C=\dfrac{2\sin 2C\sin C}{\sin(2C+C)}=\dfrac{2\sin 2C\sin C}{\sin 2C\cos C+\cos 2C\sin C}=\dfrac{2\tan 2C\tan C}{\tan 2C+\tan C}=\dfrac{4\tan C}{3-\tan^{2}C}=\dfrac{4}{\dfrac{3}{\tan C}-\tan C}$$
Given the conditions for the acute triangle, $C \in (\dfrac{\pi}{6}, \dfrac{\pi}{4})$ leads to $\tan C \in (\dfrac{\sqrt{3}}{3}, 1)$. Since $S=\dfrac{4}{\dfrac{3}{\tan C}-\tan C}$ is increasing within the domain for $\tan C$, we conclude:

$$S \in \left(\dfrac{\sqrt{3}}{2}, 2\right)$$

Therefore, the range of values for $S$ in this acute triangle is $\boxed{\left(\dfrac{\sqrt{3}}{2}, 2\right)}$.