I'm quite certain that this should be trivially simple, but it's very late and I'm not that bright at the best of times:
$\{(X_\lambda, \mathcal{U}_\lambda)\,|\,\lambda \in \Lambda\}$ is a family of topological spaces with natural projection
$pr_\mu\,:\, \prod\limits_{\lambda\in\Lambda} X_\lambda \rightarrow X_\mu, \quad (x_\lambda)_{\lambda\in\Lambda} \mapsto x_\mu.$
I need to prove that $G \subseteq \prod X_\lambda$ is open in the product topology if and only if $G = \bigcup\limits_{\alpha\in A} G_\alpha$, and each $G_\alpha = \prod\limits_{\lambda \in \Lambda} G_{\alpha \lambda}$, with each $G_{\alpha\lambda}\in \mathcal{U}_\lambda$ and $G_{\alpha\lambda} = X_\lambda$ for all but finitely many $\lambda$. The product topology is then taken to be the topology induced by $\{pr_\mu\,|\,\mu\in\Lambda\}$.
Then I need to show that each $pr_\mu$ is an open mapping.
I appreciate that this is basically just writing out the definition of the product topology, and it feels like it's very close to clicking in my brain, but I keep getting jumbled and the devil Time ruthlessly pushes ever forward.