Let I be an open interval in $\displaystyle \Re$ and let $\displaystyle f:I \rightarrow \Re$ be a differentiable function.

Let T be the set $\displaystyle T = \{(x,y) \in I \times I : x<y\} $. I can show that this is a connected subset of $\displaystyle \Re^2$ with the standard topology.

Let $\displaystyle g: T \rightarrow \Re$ be a function defined by $\displaystyle g(x,y) = (f(x) - f(y)) / (x-y)$. Prove that $\displaystyle g(T) \subset f'(I) \subset \overline{g(T)}$.

Show that $\displaystyle f'(I)$ is an interval.

This last deduction would be trivial if g(T) was a connected set, but I don't see why this should necessarily be true. (It would be true if g was a continuous function but we are not given this. Nevertheless, I think there may be a way of showing it).