Show that if $\prod_\alpha X_\alpha$ is normal then so is $X_\alpha$. 
Show that if $\prod_\alpha X_\alpha$ is normal then so is $X_\alpha$.

This a question of proof-verification.So please suggest the required edits and fault in the logic but please don't give a different proof all together.
Proof:


*

*Let $A_\alpha,B_\alpha$ be disjoint closed subsets in $X_\alpha$.
Consider the projection map $\pi_\alpha :\prod_\beta X_\beta\to X_\alpha$
.

*Then since $\pi_\alpha$ is open and continuous so $\pi_\alpha^{-1}(A_\alpha) $ and $\pi_\alpha^{-1}(B_\alpha)$ is closed in $\prod_\alpha X_\alpha$ which is normal .

*So there exists two disjoint closed sets $U,V$ in $\prod_\alpha X_\alpha$ such that $\pi_\alpha^{-1}(A_\alpha) \subseteq U$ and $\pi_\alpha^{-1}(B_\alpha)\subseteq V$.

*Then $A_\alpha \subseteq \pi(U)$
and $B_\alpha \subseteq \pi(V)$ since $\pi$ is open and surjective.Obviously $\pi(U)\cap \pi(V)=\emptyset$


Looking forward to hear from you all.

In point 2, you don't need to mention that $\pi_{\alpha}$ is open.
In point 3, you mean that $U, V$ are open.
There is a major error in point 4, because $\pi_{\alpha}(U)$ and $\pi_{\alpha}(V)$ need not be disjoint. (Also, $\pi_{\alpha}$ being open has nothing to do with the inclusions $A_{\alpha} \subseteq \pi_{\alpha}(U)$ and $B_{\alpha} \subseteq \pi_{\alpha}(V)$. Instead, this shows that $\pi_{\alpha}(U)$ and $\pi_{\alpha}(V)$ are open.)
