The key point is that since $f''(x)\neq 0$ for all $x\in (a,b)$ and $(a,b)$ is an open interval, the function $f'$ has neither a maximum, nor a minimum on $(a,b)$. In other words, for any $x\in (a,b)$, one can find $u,v\in (a,b)$ such that $f'(u)<f'(x)<f'(v)$. (This is the only way in which the assumption will be used).
It follows the the set $J:=f'((a,b))$ is an open interval. Indeed, this is an interval by the intermediate value theorem (no need to use the more subtle Darboux theorem here), and this interval is open by the above remark.
Consider the triangle $\Delta=\{ (x_1,x_2)\in (a,b)\times (a,b);\; x_1<x_2\}$, and the map $\Phi:\Delta\to \mathbb R$ defined by
$$\Phi(x_1,x_2):=\frac{f(x_2)-f(x_1)}{x_2-x_1}\cdot $$
The map $\Phi$ is continuous. Since $\Delta$ is a connected set, it follows that $I:=\Phi(\Delta)$ is a connected subset of $\mathbb R$, i.e. an interval.
Now, observe that $J=f'((a,b))$ is contained in the closure of $I=\Phi(\Delta)$. Indeed, since $(a,b)$ is right-open, we may write $f'(x)=\lim_{z\to x^+}\frac{f(z)-f(x)}{z-a}$ for any $ x\in (a,b)$, so that $f'(x)\in \overline{\Phi(\Delta)}$.
Since $J$ is an open set and $I$ is an interval, it follows that in fact $J\subset I$. So we have $$f'((a,b))\subset \Phi(\Delta)\, ,$$ which gives the required result.
Edit Note that the result may be wrong is $f'$ has a maximum or a minimum on $(a,b)$. For example consider $f(x)=\sin x$ on $(a,b)=(-\frac\pi2,\frac\pi2)$. Then $f'(0)=1$, but $f(x_2)-f(x_1)=\int_{x_1}^{x_2} \cos t\, dt< x_2-x_1$ whenever $x_1<x_2$, because $\cos t<1$ for all but one $t\in (a,b)$. More generally, the result fails if $f'$ has an isolated maximum or an isolated minimum at some point $x\in (a,b)$.
The inequality at which you arrived implies that $|f'(x) - f'(y)| \leq M$ for all $0 < x,y < 1$. If $f'$ is unbounded on $]0,1[$, then for every $B > 0$ there are some $0 < x,y < 1$ such that $|f'(x) - f'(y)| > B$; hence $f'$ is bounded on $]0,1[$ from above by $M$. Then mean-value theorem gives that for every pair of $0 < x,y < 1$ there is some $x < \xi < y$ such that $|f(x)-f(y)| \leq |f'(\xi)||x-y| \leq M|x-y|$. If $\varepsilon > 0$, the number $\delta := \varepsilon/M$ is such that $|x-y| < \delta$ implies $M|x-y| < \varepsilon$; hence $f$ is uniformly continuous on $]0,1[$.
Best Answer
As you suspect, your argument is flawed. Here is a simpler way:
$f(c) > 0$, so by the Mean Value Theorem, there exist $\alpha \in (a,c)$ and $\beta \in (c,b)$ such that $f'(\alpha) > 0$ and $f'(\beta) < 0$.
Now use the Mean Value Theorem on the function $f'$ and the interval $(\alpha,\beta)$.