I was re reading rudin and came across a familiar theorem. I thought of a different proof but I'm unsure if it's correct.
Th. If $f$ is a continuous mapping from metric space $X$ to $Y$ and if $E$ is a connected subset if $X$ then $f(E)$ is connected.
My proof attempt: Assume $f(E)$ disconnected $\implies f(E)=A \cup B$ for disjoint open nonempty sets $A,B \in Y$. Consider the preimages
$f^{-1}(f(E))=E=f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$
But since $f$ is continuous the preimage of open sets $A$ and $B$ is open in $X$.
So we get $E$ as the union of two nonempty open sets. It's intersection must be empty because if $x\in f^{-1}(A)\cap f^{-1}(B)$ then $f(x)\in A\cap B$ but this was chosen to be empty.
So E is disconnected which is a contradiction.