$A\times B$ connected component implies $A,B$ connect components $X,Y$ topological spaces. I want to show that if $C\subseteq X\times Y$ is a connected component, then $C=A\times B$ where $A,B$ are connected components of $X,Y$.
What I have so far, is that any continuous $f:\;C\to\{0,1\}$ must be constant - so pick $f_A:\;A\to\{0,1\}$ and $f_B:\;B\to\{0,1\}$ both continuous. What I'd like to do is somehow compose them to get a continuous map $F:A\times B\to\{0,1\}$ which will be constant, and then argue that $f_A,f_B$ must have been constant, so $A,B$ connected (components?).
First, would this idea work? If so, any ideas on how I can mix up $f_A,\;f_B?$

One can indeed show that $A, B$ connected implies $A \times B$ connected. There are several answers on this site that address this. 
So if $C$ is a connected component of $X \times Y$, then consider the continuous projections $p_X,p_Y$ onto the component spaces. Continuous images preserve connectedness so $A = p_X[C], B = p_Y[C]$ are connected and clearly $C \subseteq A \times B$. The latter set is connected by the first lemma and $C$ is a maximal connected set (definition of a connected component), hence equality $C = A \times B$ has been shown.
$A$ must be a component of $X$, or else there would be strictly larger connected $A \subset A' \subseteq X$, but then $A' \times B$ would also be connected and bigger than $C$, which cannot be by maximality of $C = A \times B$. Similar reasoning applies to $B$. So both $A$ and $B$ are components of $X$ resp. $Y$. 
The proof of the first mentioned lemma could proceed by starting with a continuous function $f: A \times B \rightarrow \{0,1\}$ and showing it is constant, using that $f$ restricted to all sets of the form $\{x\} \times B$ and $A \times \{y\}$ are constant (these sets are connected, being homeomorphic to $B$ resp. $A$, which are connected).
