Hi.....
Let f:A------ R
is a continuous function and A is a connected set
prove that:
{ (x,f(x)):x in A}
is a connected set
You want to prove that $\displaystyle Z = f(A) $ is connected. Consider a continuous surjective map $\displaystyle l: A \to Z $. Let $\displaystyle Z = X \cup Y $ be a separation of $\displaystyle Z $ into 2 disjoint nonempty sets open in $\displaystyle Z $. Then $\displaystyle l^{-1}(X) $ and $\displaystyle l^{-1}(Y) $ are disjoint sets whose union is $\displaystyle A $. They are open and form a separation of $\displaystyle A $. Contradiction.
The proof for this is very similar to what particlejohn suggested. Let $\displaystyle G = \{ (x,f(x)):x \in A\}$ and let G=X∪Y be a partition of G into open sets X and Y. Let $\displaystyle U = \{x\in A: (x,f(x))\in X\}$ and let $\displaystyle V = \{x\in A: (x,f(x))\in Y\}$. Then U, V are open and disjoint, and their union is A ... .