Proposition 4.12.3. Let $X$ be a connected space, $Y$ be a topological space, and $f \in C(X; Y)$, then $f(X)$ is connected.
Proof. Let $U, V \subset Y$ be open with $U \cap f(X), V \cap f(X) \ne \emptyset$. By continuity of $f$ and connectedness of $X$, $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) \ne \emptyset$. Hence $U \cap V \ne \emptyset$.$\square$