= Solution
{c}
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. Hence
$$
\boxed{f(A)=\{n_0\}.}
$$
Equivalently 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.
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