If each $X_n$ is countably compact then so is $\displaystyle\prod_{n\in\mathbb{N}}X_n$. 
Let be $X_n$ countably compact and first countable for every $n\in \mathbb{N}$. Then  $\displaystyle\prod_{n\in\mathbb{N}}X_n$ is countably compact.

I have proved some equivalences. 


*

*A space $X$ is countably compact if and only if every sequence has a cluster (accumulation) point.

*Therefore, every space sequentially compact is countably compact.

*Particularly, if $X$ is first countable, $X$ is countably compact if and only if $X$ is sequentially compact.


I don't know if this could be useful by the way.
If we take a sequence $\{(x_1^k,x_2^k,...)\}_{k\in\mathbb{N}}\subseteq \displaystyle\prod_{n\in\mathbb{N}}X_n$, I would know every projection sequence has a convergent subsequence. But I don't think we can construct a convergent subsequence of $\{(x_1^k,x_2^k,...)\}_{k\in\mathbb{N}}$ from here.
Any hint?
 A: You actually can do it that way, with a bit of trickery. Let $\sigma=\langle x^{(n)}:n\in\Bbb N\rangle$ be a sequence in the product. where $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb N\rangle$. There are an infinite $N_0\subseteq\Bbb N$ and an $x_0\in X_0$ such that the sequence $\langle x_0^{(n)}:n\in N_0\rangle$ converges to $x_0$ in $X_0$. Suppose that $m\in\Bbb N$, $N_n$ is an infinite subset of $\Bbb N$, and $\langle x_m^{(n)}:n\in N_m\rangle$ converges to some $x_m\in X_m$; then there are an infinite $N_{m+1}\subseteq N_m$ and an $x_{m+1}\in X_{m+1}$ such that $\langle x_{m+1}^{(n)}:n\in N_{m+1}\rangle$ converges to $x_{m+1}$ in $X_{m+1}$. In this way we can recursively choose infinite subsets $N_k\subseteq\Bbb N$ and points $x_k\in X_k$ such that for each $k\in\Bbb N$, $N_{k+1}\subseteq N_k$ and $\langle x_k^{(n)}:n\in N_k\rangle$ converges to $x_k$.
Let $x=\langle x_k:k\in\Bbb N\rangle$; we want to find a subsequence of $\sigma$ converging to $x$. Let $n_0=\min N_0$. Given $n_k\in N_k$, let $n_{k+1}=\min(N_{k+1}\setminus\{n_k\})$. You can easily verify that $\langle n_k:k\in\Bbb N\rangle$ is strictly increasing, so that $\sigma'=\langle x^{(n_k)}:k\in\Bbb N\rangle$ is a subsequence of $\sigma$. I’ll leave it to you to verify that $\sigma'$ converges to $x$; use the facts that any basic open nbhd of $x$ in the product restricts only finitely many coordinates, and that $n_k\in\bigcap_{\ell\le k}N_\ell$ for each $k\in\Bbb N$. Leave a question if you get stuck. 
