$(1)$ To prove: By the Law of Sines, we know that

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$

where $R$ is the radius of the circumcircle.

We can then express $a$, $b$, and $c$ as

$$a = 2R\sin A, \quad b = 2R\sin B, \quad c = 2R\sin C.$$

Given that $a-b=b\cos C$, we have

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

Since $A=\pi-(B+C)$, it follows that $\sin A=\sin (B+C)=\sin B\cos C+\cos B \sin C$, which gives us

$$\sin B\cos C + \cos B\sin C - \sin B = \sin B\cos C.$$

After simplifying, we get

$$\cos B\sin C = \sin B,$$

which implies that

$$\tan B = \frac{\sin B}{\cos B}.$$

Therefore, we can conclude that

$$\boxed{\sin C = \tan B}.$$

$(2)$ By the Law of Cosines, we have:

$$c^2 = a^2 + b^2 - 2ab\cos C = a^2 + b^2 - 2a(a-b) = b^2 + 2b - 1 = (b+1)^2 - 2.$$

From $a-b=b\cos C$, we find $b$ as

$$b = \frac{a}{1+\cos C} = \frac{1}{1+\cos C}.$$

As $C$ is acute, $\cos C$ is between $0$ and $1$, so

$$\frac{1}{2} < b < 1.$$

Let $f(b) = (b+1)^2 - 2$. This function is monotonically increasing in the interval $\left(\frac{1}{2}, 1\right)$.

Thus, $f(b)$ is in the interval $\left(\frac{1}{4}, 2\right)$, which implies

$$\frac{1}{2} < c < \sqrt{2}.$$

Hence, the range of values for $c$ is

$$\boxed{\left(\frac{1}{2}, \sqrt{2}\right)}.$$