### Proof of $(Ⅰ)$:

Given the equation $a^2=\frac{1-\cos A}{1-\cos B}$, we can apply the double angle cosine formula to transform it into:
$$a^{2}=\frac{1-\cos^2 A}{1-\cos^2 B} = \frac{\sin^{2}A}{\sin^{2}B}$$
Using the half-angle formula, this becomes:
$$a^{2}=\frac{\sin^{2}\frac{A}{2}}{\sin^{2}\frac{B}{2}}$$

Given that $A$, $B \in (0,\pi)$, it follows that $\sin \frac{A}{2} > 0$ and $\sin \frac{B}{2} > 0$. Thus, we can simplify the equation to:
$$a=\frac{\sin\frac{A}{2}}{\sin\frac{B}{2}}$$

With $b=1$, we have:
$$a\sin \frac{B}{2}=b\sin \frac{A}{2}$$
This leads to:
$$2\sin \frac{A}{2}\cos \frac{A}{2}\sin \frac{B}{2}=2\sin \frac{B}{2}\cos \frac{B}{2}\sin \frac{A}{2}$$
Which simplifies to:
$$\cos \frac{A}{2}=\cos \frac{B}{2}$$

Given that $\frac{A}{2}, \frac{B}{2} \in (0,\frac{\pi}{2})$, it implies that:
$$\frac{A}{2}=\frac{B}{2} \Rightarrow A=B$$

This contradicts the assumption that $\triangle ABC$ is non-isosceles. Therefore, there does not exist a triangle $\triangle ABC$ that satisfies the given condition.

### Calculation for $(Ⅱ)$:

Given $a^2=\frac{1-\cos A}{1-\cos B}$, we can rewrite it as:
$$a^{2}\left(1-\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\right)=1-\frac{b^{2}+c^{2}-a^{2}}{2bc}$$
This simplifies to:
$$a^{2}\frac{(b+a-c)(b-a+c)}{2ac}=\frac{(b+a-c)(a-b+c)}{2bc}$$

Since $b+a-c > 0$, we can further simplify to:
$$a^2\frac{(b-a+c)}{2ac}=\frac{(a-b+c)}{2bc} \Rightarrow \frac{1}{c}=\frac{ab-1}{ab+1}(a-b)$$

Therefore:
$$\frac{1}{c}\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{ab-1}{ab(ab+1)}$$

Let $ab-1=t$. If $t=0$, then:
$$\frac{1}{c}\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{ab-1}{ab(ab+1)}=0$$

If $t \neq 0$, then:
$$\frac{1}{c}\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{t}{(t+1)(t+2)}=\frac{1}{t+\frac{2}{t}+3}$$

For $t > 0$, we have:
$$\frac{1}{c}\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{1}{t+\frac{2}{t}+3} \leq \frac{1}{3+2\sqrt{2}}=3-2\sqrt{2}$$
with equality at $t=\sqrt{2}$, i.e., $ab=1+\sqrt{2}$.

For $-1 < t < 0$, $t+\frac{2}{t} \in (-\infty ,-3)$, $t+\frac{2}{t}+3 < 0$, which does not satisfy the requirement.

Hence, the maximum value of $\frac{1}{c}\left(\frac{1}{b}-\frac{1}{a}\right)$ is $\boxed{3-2\sqrt{2}}$.