let $A\subseteq \mathbb{R}^m$ be an open set, $f:A\rightarrow \mathbb{R}^n$ a differentiable function. If $a,b\in A$ are such that \begin{align*} \left[a,b\right]=\left\{\left(1-t\right)a+tb\::\:t\in \left[0,1\right]\right\}\subseteq A \end{align*} We define \begin{align*} \begin{split} (a,b) &= \{(1-t)a+tb : t\in (0,1)\} \subseteq A \\ &= [a,b]\setminus\{a,b\} \subseteq A \end{split} \end{align*}
then there exists $c\in \left(a,b\right)$ such that $f\left(b\right)-f\left(a\right)=Df\left(c\right)\cdot \left(b-a\right)$
I must comment that the professor has rewritten the theorem, therefore the writing of the book and what is proposed here are not written in the same way, therefore I made a demonstration adjusted to the rewriting of the theorem and what interests me at to ask the question here is to know if the proof that I have developed is correct and if it is not, I help with suggestions and/or help to prove the theorem correctly.
Proof:
Let $b=a+h$ then we must find a point $c\in(0,1)$ defined as $c=(1-a)t_0-bt_0=a+t_0h$, with $t_0\in(0,1)$ such that:
\begin{align*} f(b)−f(a)=Df(c)(b−a) \end{align*}
Note: $h=b-a\quad\rightarrow \quad f(a+h)−f(a)=Df(c)h$
Let $\alpha:[0,1]\rightarrow\mathbb R$ be defined by $\alpha(t)=(1-t)a-ta$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$. Also let $g(t)=(f\circ\alpha)(t)$, as the composition of differentiable functions is differentiable then $g$ is differentiable.
now by the definition of derivative of the composition of functions we have: \begin{align*} g'(t)=Df(\alpha(t))\cdot \alpha'(t)\tag1 \end{align*} where $\alpha'(t)=(b-a)$ and $f(\alpha(t))=f((1-t)a+tb)$ then $(1)$ is written as:
\begin{align*} g'(t)=Df((1-t)a+tb)\cdot(b-a)&=Df(a+th)\cdot h \tag2 \end{align*}
The ordinary mean-values theorem implies that
\begin{align*} g'(t_0)=\frac{g(1)-g(0)}{1-0}\\ g(1)-g(0) = Df(t_0)\cdot 1 \tag3 \end{align*}
for some $t_0\in(0,1)$. The equation $(3)$ can be rewritten in the form \begin{align*} f(a+h)-f(a)=Df(a+t_0h)\cdot h\\ f(b)-f(a)=Df(c)\cdot h \\ \quad\hfill\square \end{align*}
Sadly, we can't really help you prove the theorem correctly, since the claim is false (check the paragraph starting with "The reason why there is no analog of mean value equality is the following: ..."): https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorem_for_vector-valued_functions
The part where your argument breaks down is when you try to apply the "ordinary mean value theorem" on $g$ : assuming $n \neq 1$, $g$ does not take its values in $\mathbb{R}$, since $f$ has its values in $\mathbb{R}^n$.