Product of connected sets is connected

Since the 'product of subspaces = subspace of products' (see this), the question boils down to showing that the product of two connected spaces is connected. The answer to that question, using the OP's hints, can be found here.

Yes, you just need the definition of homeomorphisms to prove the statement. I assume you know this fact 1: Let $S,T$, $S subset T$ two subset of a topological space $X$, assume that $S$ is connected and $T$ is not connected. If $(U,V)$ is a separation of $T$ then $S subset U$ or $S in U$. Here is a proof of the following fact2: Let X and Y two topological spaces, then $X times Y$ is connected $Leftrightarrow X,Y$ are connected. $(Leftarrow)$ suppose that $X,Y$ are connected and $(U,V)$ is a couple of open nonempty sets which unconnect $X times Y$ which means $X times Y= U bigcup V$. Now let $(x_0,y_0) in U$. The set ${x_0} times Y$ is connected because it is homeomorphic to $Y$(it is very easy to show!). ${x_0}times Y$ contains $(x_0,y_0)$ then for fact 1 ${x_0} times Y subset U$. On the other hand $forall y in Y$, the set $X times y$ is connected because is homeomorphic to X and contains $(x_0,y) in U$. Again for fact 1 we can deduct that $X times y subset U$. Then $X times Y= bigcup_{y in Y} X times y subset U$ $Rightarrow$ $X times Y=U$ and $V$ is empty, which is absurd because we supposed that $V$ is nonempty. This contradiction shows that $X times Y$ can't be unconnected then is connected. $(Rightarrow)$ Assume that $X times Y$ is connected, define the canonical projection $pi: X times Y rightarrow X$, $pi$ is surjective and a continuous map, and for the principal theorem of connection you easily conclude that $X$ is connected. The same reasoning shows that also $Y$ is connected. $Box$ The principal theorem of connection says that if $X,Y$ are a topological spaces with X connected and $f: X rightarrow Y$ is a continuous map, then $f(X)$ is connected.