I am proving that if the product space $\prod_{\alpha \in J} X_\alpha$ is normal, then any $X_\alpha$ is normal.
So far I have this:
Let $A$ be closed in $X_\alpha$, then $\pi_\alpha^{-1}(A)$ is a closed set of $\prod_{\alpha \in J} X_\alpha$. Given some neighborhood $U$ of $A$, then $\pi_\alpha^{-1}(U)$ is an open set that contains $\pi_\alpha^{-1}(A)$. Since the product space is normal, there exists some open set $V\subset \pi_\alpha^{-1}(U)$ s.t. $\pi^{-1}_\alpha(A)\subset V$ and $\overline{V} \subset \pi_\alpha^{-1}(U)$. Then $\pi_\alpha (V)$ is an open set (since $\pi_\alpha$ is an open map), $A\subset \pi_\alpha(V) \subset U$ and $\overline{\pi_\alpha(V)} \subset \pi_\alpha(\overline{V}) \subset U$.
Is my thought correct? Any comments?
Suppose some factor is not normal. Choose two disjoint closed sets that cannot be separated by open sets.
Then consider closed sets in the product that use these closed sets at the relevant factor, and the whole space for each other factor. These also cannot be separated by open sets, proving the product is not normal.