# 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.