On proving that if $\prod_{\alpha \in \Lambda}X_\alpha$ is $T_i$ with $i \in \{3,4\}$ then $X_\alpha$ is $T_i$ for all $\alpha$. As the title sates, I want to prove that if On proving that if $X = \prod_{\alpha \in \Lambda}X_\alpha$ is a $T_3$ or $T_4$ space, then each space $X_\alpha$ is $T_3$ or $T_4$ respectively.
In the case for when $X$ is $T_3$, I have proceeded as follows: let $\beta \in \Lambda$, and $x \in F^c$ with $F$ a closed set in $X_\beta$. Now, let $(x_\alpha)_\alpha$ be an element of the product such that $x_\beta  = x$. By regularity of the product space, since $F \times \prod_{\alpha \neq \beta} X_\alpha$ is closed, there exists $U$ open so that 
$$
(x_\alpha)_{\alpha \in \Lambda} \in U \subseteq \overline{U} \subseteq F^c \times \prod_{\alpha \neq \beta}X_\alpha.
$$
Moreover, we can take $U$ basic so that $U = \prod_{\alpha} U_\alpha$ with each $U_\alpha$ open and only finite being proper subsets. Finally, by applying the projection to $X_\beta$, we get
$$
x \in \pi_\beta(U) \subseteq\pi_\beta(\overline{U}) \subseteq F^c
$$
and since $\overline{\prod_\alpha U_\alpha} = \prod_\alpha\overline{U_\alpha}$, we have
$$
x \in U_\alpha \subseteq \overline{U_\alpha} \subseteq F^c
$$
as desired.
Now, I wanted to know whether the approach can be extended to the other case. I presume it will not be immediate (if even possible) because I am strongly using that $U$ can be taken as a 'box like' open set and so its closure behaves well with respect to the projection. Moreover, since I have not used that $X$ is $T_1$, I am also wondering whether $X$ being normal implies each space to be normal, i.e. if we can drop the hypothesis in both cases. Any ideas?
 A: If you know that the $X_\alpha$ are all $T_1$, then it's easy: fix any point $p \in \prod_\alpha X_\alpha$ and define for a fixed $\alpha \in \Lambda$:
$$e_\alpha: X_\alpha \to \prod_{\beta \in \Lambda} X_\beta \text{ by: } e_\alpha(x)_\beta = x \text{ if } \beta=\alpha \text{ and } e_\alpha(x)_\beta = p_\beta \text{ if } \beta \neq \alpha$$
Then $e_\alpha$ is continuous by the universal property of maps into products: $\pi_\beta \circ e_\alpha$ is either the identity on $X_\alpha$ for $\beta=\alpha$ or the constant map with value $p_\beta$ for all other $\beta$; always continuous.
$\pi_\alpha$ is a continuous inverse from $e_\alpha[X_\alpha]$ back to $X_\alpha$, showing $e_\alpha$ to be an embedding of $X_\alpha$ into the large product.
As any subspace of a regular space is regular and any closed subspace of  a normal space is normal, we are done with the implication. We only use $T_1$-ness for the normal case. Note that $e_\alpha[X_\alpha] = \prod_{\beta \in \Lambda} C_\alpha$, where $C_\beta = X_\alpha$ if $\beta = \alpha$, and otherwise $C_\beta = \{p_\beta\}$, a product of closed subspaces, hence closed in the product.
(so we only need one closed singleton in each factor space, if we're being pedantic).
Another direct proof for normality without $T_1$-ness can be ( I think) achieved by separating $F,G \subseteq X_\alpha$ using the closed box subsets of the product by using $X_\beta$ in all other factors and using that projections are open maps. Try it out.
A: The product of two Sorgenfrey spaces which are
normal (lower limit topology for R), is not normal.  
If X is not empty and all the spaces are T1, then
each $X_a$ is homeomorphic to a closed subspace
of X, hence normal.
