The following problem is 'absolutely' clear:
Problem: Let $f$ be continuous on the interval $[a,b]$ and $n$-times differentiable on $(a,b)$ and $f$ vanishes on $n+1$ points $x_0< x_1 < \dots <x_n$ in $[a,b]$. Then there exists an $\xi \in [a,b]$ such that $f^{(n)}(\xi)=0$
My proof: Induction over $n$.
Let $n=1$. Then $f$ is one times differentiable and vanishes on two distinct points $x_0<x_1$ in $[a,b]$ that means $f(x_0)=f(x_1)=0$, so applying Rolle's theorem to this situation there will be a $\xi \in (x_0,x_1)$ such that $f'(\xi)=0$.
Note: As far as I am concerned it is fair to say that $a \leq x_0 < \xi < x_1 \leq b$ so $\xi$ will not be $a$ nor $b$. It might still be true to say that $\xi \in [a,b]$ or the problem above isn't correctly written.
Let $n \geq 1$. Here I wouldn't know what to 'rigorously' write down to make this proof complete. It is absolutely clear what happens to me. By successively applying Rolle's theorem over and over again to the situation at hand, namely: $$0=f(x_0)=f(x_1) = \dots =f(x_n)=0 \\\text{with } x_0<x_1 < \dots< x_i< \dots < x_n$$ We will always find roots between the already given ones for the next higher derivative by successfully applying Rolle's theorem to each of them, also in each step we will end up with minus one root then in the previous one because they're sandwiched in between the given roots and finally end up with the desired property because we did start with $(n+1)$ roots which is crucial.
While this argumentation sure is correct I can't see the typical induction character in it. How could I improve it? Or is my argumentative induction good enough? It has the character of 'counting' steps which is induction on $\mathbb{N}$. Nevertheless I am sure it's not good enough.
How about something like this:
Base case: If $f$ vanishes at both $x_0<x_1$ for $x_0, x_1 \in [a,b]$ then there is some $c \in [a,b]$ so that $f'(c)=0$.
You give a nice proof of this already above.
Inductive step: Suppose the statement " If $f$ if $n$ times differentiable and vanishes at $x_0<x_1<...<x_n$ for $x_i \in [a,b]$, then there exists some $c \in [a,b]$ so that $f^{(n)}(c)=0$" is true for $n$. Now we need to show it for $n+1$.
Then we have $x_0<...<x_n<x_{n+1}$ and some function $g$ that is $n+1$-times differentiable on $[a,b]$. Now, as you said, we apply Rolle's Theorem to each interval $[x_i, x_{i+1}]$ and conclude that there is some $c_i \in [x_i, x_{i+1}]$ so that $g'(c_i)=0$. So now lets look at the function $g'$. It is $n$-times differentiable and vanishes on $c_0<c_1<...<c_n$. Thus, it satisfies the inductive assumption and so there is some $c \in [a,b]$ so that ${g'}^{(n)}(c)=0$. But this means $g^{(n+1)}(c)=0$ and so proves the result for $n+1$.