 Solution

Since $A$ is nonempty, choose $a_0\in A$ and put $n_0=f(a_0)$. The fiber $U=f^{-1}(\{n_0\})$ is nonempty and open by part (c). Its complement is the union of the other integer fibers and is therefore also open. If the image contained a second value, that complement would be nonempty, and these two disjoint open subsets would separate $A$. connectedness excludes this. HenceEquivalently a continuous function from a connected space to the discrete space of integers is constant. The proof uses relative openness and does not require the original subset $A$ to be open in the plane.