Proof. Assume, on the contrary, that \(f(E)\) is not connected i.e \(f(E) = A\cup B\) where \(A\) and \(B\) are non-empty, separated subsets of \(Y\). Putting \(G = E \cap f^{-1}(A)\) and \(H = E\cap f^{-1}(B)\), we have \(E = G \cup H\), and neither \(G\) nor \(H\) is empty.
Since \(A\subset \overline {A}\,,\,\) we have \(G\subset f^{-1}\big (\overline {A}\big )\). \(f^{-1}\big (\overline {A}\big )\) is closed since \(f\) is continuous. Hence \(\overline {G} \subset f^{-1}\big (\overline {A}\big )\). Thus, \(f\big (\overline {G}\big ) \subset \overline {A}\). Since \(f(H) = B\) and \(\overline {A} \cap B = \emptyset \), we conclude that \(\overline {G}\cap H = \emptyset \).
The same argument shows that \(\overline {H}\cap G = \emptyset \). Thus \(G\) and \(H\) are separated. This contradicts that \(E\) is connected. Therefore \(f(E)\) is connected.


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