(Ⅰ) Proof:
Given that $\cos A=3\cos B-4\cos^3 B$, this can be rewritten as:
\begin{align*}
\cos A &= 3\cos B - 4\cos^3 B \\
&= \cos B (3 - 4\cos^2 B) \\
&= \cos B \left(3 - 4 \times \frac{1+\cos 2B}{2}\right) \\
&= \cos B - 2\cos B \cos 2B \\
\cos A + 2\cos B \cos 2B &= \cos B
\end{align*}

Since $C = 2B$, we have $A = \pi - B - C = \pi - 3B$, thus we can rewrite the equation as:
\begin{align*}
-\cos 3B + 2\cos B \cos C &= \cos B \\
2\cos B \cos C - \cos B &= \cos 3B \\
2\cos B \cos C - \cos B &= \cos (B + C) \\
&= \cos B \cos C - \sin B \sin C
\end{align*}

Rearranging terms, we have:
\begin{align*}
\cos B \cos C - \cos B &= -\sin B \sin C \\
\cos B (\cos C - 1) &= -\sin B \sin C
\end{align*}

Since $\cos C + \sin B \sin C = \cos B$ and using the identity $\cos(C-B) = \cos B$, we obtain:
$$\cos(2B - B) = \cos B$$
This holds true, therefore proving that $\cos A = 3\cos B - 4\cos^3 B$.

(Ⅱ) **To find angle $B$**:
In triangle $ABC$, given $S = \frac{b^2 + c^2 - a^2}{4}$, we have:
$$\frac{1}{2} bc \sin A = \frac{b^2 + c^2 - a^2}{4}$$
This leads to:
$$\frac{1}{2} bc \sin A = \frac{1}{2} bc \cos A$$
Thus:
$$\tan A = 1$$
Therefore, we get:
$$A = 45^\circ$$
Since $b \sin B - c \sin C = a$, we have:
$$\sin^2 B - \sin^2 C = \frac{\sqrt{2}}{2}$$
Equivalently:
$$\cos 2C - \cos 2B = \frac{\sqrt{2}}{2}$$
This can be written as:
$$\cos(270^\circ - 2B) - \cos 2B = \frac{\sqrt{2}}{2}$$
Simplifying further, we get:
$$-\sin 2B - \cos 2B = \frac{\sqrt{2}}{2}$$
Hence:
$$\sin(2B + 45^\circ) = -\frac{1}{2}$$
As a result, we have two possibilities for the equation:
$$2B + 45^\circ = 210^\circ \quad \text{or} \quad 2B + 45^\circ = 330^\circ$$
Solving for $B$ yields:
$$B = 77.5^\circ \quad \text{or} \quad B = 142.5^\circ$$
However, we discard the second value because it is not possible in this context, thus the final answer is:
$$B = \boxed{77.5^\circ}$$