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.

David's reply shows why your approach does not quite work, so I will go ahead and give a hint:
Let $\beta$ be given. We want to show $X_\beta$ is normal. 
Note we must assume each $X_\alpha$, $\alpha\neq \beta$, is nonempty for the theorem to hold. For if one of them is empty, the product is empty and thus trivially normal, while $X_\beta$ could be anything.
For each $\alpha\neq\beta$ let $x_\alpha\in X_\alpha$. Let $$X^*=\{f\in \prod X_\alpha:f(\alpha)=x_\alpha\text{ for }\alpha\neq \beta\}.$$ What can you say about $X^*$?
