Pretty much any non-trivial product of uncountably many first countable spaces fails to be first countable:
Let $A$ be an uncountable index set, and for each $\alpha\in A$ let $X_\alpha$ be a first countable space with points $p_\alpha$ and $q_\alpha$ such that $p_\alpha$ has an open nbhd $U_\alpha$ such that $q_\alpha\notin U_\alpha$. Let $X=\prod_{\alpha\in A}X_\alpha$, and let $p=\langle p_\alpha:\alpha\in A\rangle\in X$; then $X$ is not first countable at $p$.
Let $\{V_n:n\in\Bbb N\}$ be any countable local base at $p$ in $X$. For each $n\in\Bbb N$ there are a finite $F_n\subseteq A$ and open sets $U(n,\alpha)$ in $X_\alpha$ for each $\alpha\in F_n$ such that the basic open set
$$B_n=\{x\in X:x_\alpha\in U(n,\alpha)\text{ for each }\alpha\in F_n\}$$
in the product $X$ is a nbhd of $p$ contained in $V_n$. Clearly $\{B_n:n\in\Bbb N\}$ is a local base at $p$.
Now let $C=\bigcup_{n\in\Bbb N}F_n$; each $F_n$ is finite, so $C$ is countable. $A$ is uncountable, so there is an $\alpha_0\in A\setminus C$. Let
$$W=\{x\in X:x_{\alpha_0}\in U_{\alpha_0}\}\;,$$
and note that $W$ is an open nbhd of $p$. Let $z=\langle z_\alpha:\alpha\in A\rangle\in X$ be defined by
$$z_\alpha=\begin{cases}
p_\alpha,&\text{if }\alpha\in C\\
q_\alpha,&\text{if }\alpha\in A\setminus C\;;
\end{cases}$$
then $z\in B_n\setminus W$ for each $n\in\Bbb N$, so for all $n\in\Bbb N$ we have $B_n\nsubseteq W$, and $\{B_n:n\in\Bbb N\}$ therefore cannot be a local base at $p$ after all.
One can prove a somewhat similar result for box products of countably infinitely many factors.
For each $n\in\Bbb N$ let $X_n$ be a first countable space with a non-isolated point $p_n$. Let $X$ be the box product of the spaces $X_n$, and let $p=\langle p_n:n\in\Bbb N\rangle\in X$; then $X$ is not first countable at $p$.
Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable local base at $p$. Without loss of generality we may assume that for each $n,k\in\Bbb N$ there are open nbhds $U(n,k)$ of $p_k$ in $X_k$ such that
$$B_n=\prod_{k\in\Bbb N}U(n,k)\;.$$
For each $k\in\Bbb N$ let $V_k$ be an open nbhd of $p_k$ such that $V_k\subsetneqq U(k,k)$; this is possible because $p_k$ is not isolated in $X_k$.
Let $V=\prod_{k\in\Bbb N}V_k$, and suppose that $B_n\subseteq V$ for some $n\in\Bbb N$. Then $U(n,k)\subseteq V_k$ for each $k\in\Bbb N$, and in particular $U(n,n)\subseteq V_n\subsetneqq U(n,n)$, which is absurd. Thus, $V$ is an open nbhd of $p$ that does not contain any member of $\mathscr{B}$, contradicting the assumption that $\mathscr{B}$ was a local base at $p$. It follows that $X$ cannot be first countable at $p$.
Your formula makes no sense because of the two conflicting $i$ indices.
We can correct it as follows (for arbitrary index set):
$$\prod_{i \in I} X_i \setminus \prod_{i \in I} C_i = \bigcup_{j\in I} \prod_{i \in I} O(i,j)$$ where $O(i,j) = X_i$ for $i \neq j$ and $O(i,j) = X_j \setminus C_j$ for $i=j$. We could write that product of $O(i,j)$ for fixed $j$ as $\pi_j^{-1}[X_j \setminus C_j]$ as well.
This holds as $f \notin \prod_{i \in I} C_i$ iff there exists some $j$ such that $f(j) \notin C_j$.
At any rate, the complement of $\prod_i C_i$ is writeable as a union of products of (sub)basic open sets and hence open.
Best Answer
What's wrong is the step where you say $X\setminus\prod_{i=1}^\infty C_i=\prod_{i=1}^\infty(X_i\setminus C_i)$. Not so.
For example, suppose $X_i=\{0,1\}$ and $C_i=\{0\}$. What is $X\setminus\prod_{i=1}^\infty C_i$ and what is $\prod_{i=1}^\infty(X_i\setminus C_i)$?
Hint: $\prod_{i=1}^\infty C_i=\bigcap_{i=1}^\infty\pi_i^{-1}(C_i)$.
Use the facts that the projection maps are continuous, and continuous counterimages of closed sets are closed, and intersections of closed sets are closed.
By the way, "countable" has nothing to do with it; an arbitrary product of closed sets is closed.