I developed a proof of the following statement but I'm not sure if it is correct:
Proposition: Let $\lbrace X_\alpha \rbrace_{\alpha \in I}$ a collection of topological spaces and consider $X := \prod_{\alpha \in I} X_\alpha$. Then if $X$ is Hausdorff, then each $X_\alpha$ is Hausdorff. Same statement if $X$ is regular or normal.
"Proof": Let $(a_\alpha)_{\alpha \in I} \in X$ and define for each $\beta\in I$ the subspaces $Y^\beta_\alpha = \lbrace a_\alpha \rbrace$ if $\alpha \neq \beta$ and $Y^\beta_\alpha = X_\beta$ if $\alpha = \beta$. Then for each $\beta \in I$ we have the following homeomorphism relation:
$$X_\beta \cong \prod_{\alpha \in I} Y^\beta_\alpha \subset X.$$
Then, if $X$ is Hausdorff and $X_\beta$ is homeomorphic to a subspace of a Hausdorff space, I conclude that $X_\beta$ is Hausdorff and same if $X$ is regular.
Now for normality, is a fact that if $X$ is normal, then $I$ is countable or finite. But we already know that if $I$ is countable or finite, then is metrizable and the subspace $\prod_{\alpha \in I} Y^\beta_\alpha$ is also metrizable. As all metrizable space is normal, then each $X_\beta$ is homeomorphic to a metrizable subspace of $X$, then $X_\beta$ is metrizable and finally, it is normal. $_\square$
I'm not sure about the homeomorphism relations, can you see any holes in my argument?