The Stacks project

Proof. Let $A \subset f(E)$ an open and closed subset of $f(E)$. Because $f$ is continuous, $f^{-1}(A)$ is an open and closed subset of $E$. As $E$ is connected, $f^{-1}(A) = \emptyset $ or $f^{-1}(A) = E$. However, $A\subset f(E)$ implies that $A = f(f^{-1}(A))$. Indeed, if $x\in f(f^{-1}(A))$ then there is $y\in f^{-1}(A)$ such that $f(y) = x$ and because $y\in f^{-1}(A)$, we have $f(y) \in A$ i.e. $x\in A$. Reciprocally, if $x\in A$, $A\subset f(E)$ implies that there is $y\in E$ such that $f(y) = x$. Therefore $y\in f^{-1}(A)$, and then $x\in f(f^{-1}(A))$. Thus $A = \emptyset $ or $A = f(E)$ proving that $f(E)$ is connected. $\square$