Theorem: If $f$ is continuous on a connected set then the image $f(S)$ is also connected in the codomain.
Proof: Suppose that $f(C)$ is not connected, i.e., disconnected. We will show that a contradiction arises.
Suppose that $f(C)$ is disconnected. Then there exists nonempty open sets $A,B \subset f(C)$ such that $A \cap B= \emptyset$ and $f(C)=A \cup B$.
Then a necessary step is to say $C=f^{−1}(A) \cup f^{−1}(B)$. I don't know how to verify this condition.
Note that $f^{−1}(A)$ and $f^{−1}(B)$ are nonempty, otherwise, A or B would be empty which cannot happen. Furthermore, both of these sets are open from the continuity of f. We claim that $f^{−1}(A) \cap f^{−1}(B)=\emptyset$. Suppose not. Then there exists an $x \in f^{−1}(A) \cap f^{−1}(B)$ and so f(x)∈A∩B which implies that A∩B≠∅ which is a contradiction.
Therefore $f^{−1}(A) \cap f^{−1}(B)=\emptyset$ and so C is a disconnected set. But this is a contradiction.
Hence the assumption that f(C) was disconnected is false. Thus, we prove that f(C) is connected.
Best Answer
If $x\in C,$ then $f(x)\in f(C)=A\cup B$ so either $f(x)\in A$ or $f(x)\in B.$ Thus either $x\in f^{-1}(A)$ or $x\in f^{-1}(B).$