 Solution

The mean value theorem states that a function continuous on $[p,q]$ and differentiable on $(p,q)$ has some $c\in (p,q)$ with $f(q)-f(p)=f^{\prime }(c)(q-p)$.

Take any $p<q$ inside the given open interval. Differentiability implies the continuity needed on $[p,q]$, and the derivative is zero throughout it. The mean value theorem therefore gives $f(q)=f(p)$. Since the two points were arbitrary, the function is constant on the interval.