Let $X,Y,Z$ be three topological spaces, let $X\times Y$ have the product topology, let $p_X:X\times Y\to X$ and $p_Y:X\times Y\to Y$ be projective maps, and let $f:Z \to X \times Y$ be a function.
I want to prove that $f$ is continuous if and only if $p_X \circ f: Z \to X$ and $p_Y \circ f:Z \to Y$ are continuous.
Given the fact that the projective maps are continuous, one direction is clear.
Now suppose $p_X \circ f$ and $p_Y \circ f$ are both continuous and let $U \times V \subseteq X \times Y$ be open. Then $U$ and $V$ are open in $X$ and $Y$ respectively, and $U \times V = p_X ^{-1}(U) \cap p_Y ^{-1}(V)$.
Thus $f^{-1}(U \times V) = f^{-1}(p_X ^{-1}(U) \cap p_Y ^{-1}(V)) = f^{-1}(p_X ^{-1}(U)) \cap f^{-1}(p_Y ^{-1}(V)) = (p_X \circ f)^{-1}(U) \cap (p_Y \circ f)^{-1}(V)$ is open in $Z$ as the intersection of two open sets. In this last step we used the continuity of $p_X \circ f$ and $p_Y \circ f$. Hence $f$ is continuous.
I am not entirely sure about this proof. I believe it contains some error, but I can't quite put my finger on it.
All remarks, corrections or improvements would be highly appreciated.
Edit: I should clarify what I mean by the product topology: Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be topological spaces. $X \times Y$ is said to have the product topology $\tau$ if $\tau = \{ U\times V : U \in \tau_X , V \in \tau_V \}$. I hope this isn't too unorthodox a definition of the product topology. It is the one I have been working with all year. I am trying to prove the statement above using this definition of the product topology.
Second edit: In my previous edit, I gave a false definition of the product topology. It should be defined as the topology obtained from the basis $\{ U\times V : U \in \tau_X , V \in \tau_V \}$. In particular, this implies that if $U$ and $V$ are open in $X$ and $Y$ respectively, then $U \times V$ is open in $X \times Y$, but the converse is not necessarily true.
There is a problem with your proof. Not every open set in $X \times Y$ is of the type $U \times V$. So start with any open set $W$ in $X \times Y$. To show that $f^{-1}(W)$ is open pick any point $z$ in it, so $f(z) \in W$. Then use the fact that there exist open sets $U,V$ such that $f(z) \in U\times V \subset W$. Can you continue?