Theorem:
Continuous image of connected set is connected.
Proof:
Let $ (X, d) $ and $ (Y, d’) $ be any two metric spaces and $ f:X\to Y $ be continuous function.
Claim: f(X) is connected.
Suppose, if possible that f(X) is not connected.
So, there exists non-empty disjoint open subsets A and B such that $ f(X)=A\cup B, \bar{A}\cap B=\phi $ and $ \bar{B} \cap A=\phi $.
$ \therefore f^{-1}(f(X))=f^{-1}(A\cup B) $
$ X=f^{-1}(A\cup B) $
$ X=f^{-1}(A)\cup f^{-1}(B) $
Also,
$ f^{-1}(A)\cap f^{-1}(B)=\phi $
Since, A and B are open subsets of Y and $ f:X\to Y $ is continuous ,
$ f^{-1}(A) $ and $ f^{-1}(B) $ are open subsets of X.
$ X=f^{-1}(A)\cup f^{-1}(B) $
$ f^{-1}(A)\cap f^{-1}(B)=\phi $
$ \therefore $ X is disconnected, which is contradiction to X is connected.
$ \therefore $ Our assumption is wrong.
$ \therefore $ f(X) is connected.
Thus, continuous image of connected set is connected.
