Proving a lemma for a dichotomy I am currently trying to prove the Rolle theorem by dichotomy. I think it is possible and I found the following property, that might be useful : 

let $f : [a,b]\to \mathbf{R} $ be a continuous function with $f(a)-f(b)=0$, then there exist $\alpha$ and $\beta$ such that $a<\alpha<\beta<b$, with $\beta-\alpha\leq\frac{b-a}{2}$ and $f(\alpha)=f(\beta)$

To prove this lemma, I thought I'd use the IVT on the following function : 
$$\forall x \in I=\left[a,\frac{a+b}{2}\right], g(x)=f\left(x-\frac{b-a}{2}\right)-f(x)$$
to prove that there exist $\alpha \in I$ such that $g(\alpha)=0$, and then maybe proving that there exist a $\beta$ in $\left[\frac{a+b}{2},b\right]$ such that $f(\alpha)=f(\beta)$
From there, the lemma would (?) follow, 
Any Help is appreciated, 
T.D
 A: $$f(x) \ge 0, f\left(\frac{a+b}{2}\right) > 0 \\ \Huge\overset{\mmlToken{mo}{⏜}}{\overline{\phantom{ABC}}\,} \\ x = a \qquad \qquad x = b$$
Without loss of generality, assume that $f(a) = f(b) = 0$.  I'll first prove the lemma for the simple case $f \ge 0$ on $[a,b]$ with $f\left(\dfrac{a+b}{2}\right) > 0$.
The auxiliary function $$g(x)=f\left(x-\frac{b-a}{2}\right)-f(x)$$ should be defined on the right-half of the interval $$I=\left[\frac{a+b}{2},b\right].$$
This enables us to see that
\begin{align}
g\left(\frac{a+b}{2}\right) &= f(a) - f\left(\frac{a+b}{2}\right) = 0 - f\left(\frac{a+b}{2}\right) < 0\text{, and} \\
g(b) &= f\left(\frac{a+b}{2}\right) - f(b) = f\left(\frac{a+b}{2}\right) - 0 > 0.
\end{align}
Therefore, there exists $\beta \in I$ such that $g(\beta) = 0$.
$$\text{i.e.} f(\alpha) - f(\beta) = 0 \text{, where } \alpha := \beta - \frac{a+b}{2}.$$
$$f\left(\frac{a+b}{2}\right) = 0 \\ \huge \overset{\mmlToken{mo}{⏜}}{\overline{\phantom{AB}}\,} \overset{\mmlToken{mo}{⏜}}{\overline{\phantom{AB}}\,} \\ x = a \qquad \qquad x = b$$
Then, prove the lemma for another special case $f\ge 0$ on $I$ with $f\left(\dfrac{a+b}{2}\right) = 0$ and that $f$ is finitely many roots on $I$.  Since the number of roots of $f$ on $I$ is finite, we can easily reduce this into the simply case: just like a subinterval $[c,d] \subseteq I$ so that $f(x) \ne 0$ for all $x \in (c,d)$.  This reduces to the first case.

Graph of $f(x) = \begin{cases}x \sin(1/x) & \text{if } x \ne 0 \\ 0 & \text{if } x = 0 \end{cases}$ (source of picture)
The remaining case is even simpler: $f$ has infinitely many crossings with the $x$-axis on interval $I$.  An argument by length should enable you to find the desired subinterval $[\alpha,\beta]$.

Edit: this answer addresses the original version of the question.
Your lemma is false.  Take $f(x) = 689x$ defined on $[0,8964]$ as a counterexample.  $f(0)f(8964) = 0 \times 8964 = 0$, but since $f$ is strictly increasing, $f$ is injective, so no distinct interior points $\alpha$ and $\beta$ in the domain can satisfy $f(\alpha) = f(\beta)$.
