This question is regarding the same problem. I wish to present my proposed solution and get feedback on my argument, and as such, I claim that it is not a duplicate. (In particular, the other asker actually knew what he was doing, so this question will probably be sufficiently different. It also includes the discussion about the box topology, which the original question lacked.)
We are given that $\{x_n\}$ is a sequence of points in the product space $\prod X_\alpha$. We are to show that this sequence converges to $x$ if and only if $\{\pi_\alpha(x_n)\}$ converges to $\pi_\alpha(x)$. Moreover, we are to determine if this holds if we equip $X = \prod X_\alpha$ with the box topology.
We first show $\implies$. Suppose, for the sake of contradiction, that $\{\pi_\alpha(x_n)\}$ does not converge to $\pi_\alpha(x)$ for a fixed $\alpha$. Then there exists an open set $U_\alpha$ containing $\pi_\alpha(x)$ such that $X_\alpha - U_\alpha$ contains infinitely many $\pi_\alpha(x_n)$. Since every projection in the product topology is continuous, $\pi^{-1}(U_\alpha)$ is open in $X$. This contradicts the assumption that $\{x_n\} \to x$, since we now have an open set of $X$ where one of the $\{x_{n_i}\}$ clearly do not converge, where $\{x_n\} = (\{x_{n_1}\},\{x_{n_2}\},\cdots)$
We proceed to show $\impliedby$. We are given that $\{\pi_\alpha(x_n)\} \to \pi_\alpha(x)$ for every $\alpha.$ Let $\prod U_\alpha$ be an open set in $X$ containing $x$. By the definition of the product topology, we know that $U_{\alpha} \ne X_\alpha$ for only finitely many $\alpha_1, \alpha_2,\cdots,\alpha_k$. Now, since we know that $\{\pi_\alpha(x_n)\} \to \pi_\alpha(x)$, it follows by the definition of convergence that there exist numbers $N_{\alpha_1},N_{\alpha_2},\cdots,N_{\alpha_k}$ such that $\pi_{\alpha_i}(x_n) \in U_{\alpha_i}$ for $i \in \{1,\cdots,k\}$ and $n \ge N_{\alpha_i}$. Letting $N = \max\{N_{\alpha_1},\cdots,N_{\alpha_k}\}$, we see that $x_n \in \prod U_\alpha$ for $n \ge N$. (Recall that for $\alpha$ not in the finite collection, it equals the whole space $X_\alpha$, where inclusion is trivial.) We conclude that $\{x_n\} \to x$
Lastly, we consider $X$ with the box topology. In $\implies$ we did not use any particular properties of the product topology, aside from the continuity of projections, which also holds for the box topology. As such, I claim that $\implies$ still holds. However, I claim that $\impliedby$ does not hold. I ask for an example that shows this, since I am not able to find any argument I'm happy with. ("Well, our proof used properties of the product topology, so..." is only acceptable for my private notes.)
