Q:

Prove that $\prod X_\alpha$ is first (second)-countable if and only if $X_\alpha$ is first (second)-countable, $\forall \alpha\in I.$

Let $I\neq\emptyset$ numerable and $(X_\alpha,\tau_\alpha)$ a family of topological spaces. Prove the following.

$\displaystyle\prod X_\alpha$ is first-countable if and only if $X_\alpha$ is first-countable, $\forall \alpha\in I.$

$\displaystyle\prod X_\alpha$ is second-countable if and only if $X_\alpha$ is second-countable, $\forall \alpha\in I.$

All I have are the definitions and I do not know how to proceed to do the proofs.
Could you give me the idea of the proof  for 1. and 2. please ? ? ?
Note that I am not asking for the entire proof since I know this site does not work like that.

Definition
$(X,\tau)$ is first-countable if $\forall x\in X,$ has a countable local  basis of neighborhoods.
$(X,\tau)$ is second-countable if $\tau$ has a countable basis, i.e. if $\exists\beta\subset\tau:\beta $ is countable and basis for $\tau.$

A:

2) If part. Let $\beta_{\alpha}$ be a countable basis for $\tau_{\alpha}$, then try to argue that the following set
\begin{align*}
\beta=\left\{G_{\alpha_{1}}\times\cdots\times G_{\alpha_{n}}\times\prod_{\alpha\ne\alpha_{1},...,\alpha_{n}}X_{\alpha}: G_{\alpha_{k}}\in\beta_{\alpha_{k}},~\alpha_{k}\in I,~k=1,...,n,~n\in{\bf{N}}\right\}
\end{align*}
is a countable basis for $\displaystyle\prod X_{\alpha}$.
Only if part. Let $\beta$ be a countable basis for $\displaystyle\prod X_{\alpha}$, then try to argue that the following set
\begin{align*}
\beta_{\alpha}=\{\pi_{\alpha}(G): G\in\beta\}
\end{align*}
is a countable basis for $\tau_{\alpha}$, where 
\begin{align*}
\tau_{\alpha}&:\prod X_{\alpha}\rightarrow X_{\alpha},~~~~(x_{\alpha})\rightarrow x_{\alpha}.
\end{align*}

