Is the right idea? I feel like I am missing detail somewhere. Thanks for any help.
Let $\{X_n\}_{n\in\mathbb{N}}$ be a countable collection of indiscrete spaces. Consider $\prod_{n\in\mathbb{N}} X_n$ endowed with the product topology. To show that the product topology here is just the indiscrete topology, we must show that the only open sets in the product are $\emptyset$ and $\prod_{n \in \mathbb{N}} X_n$. Let $X = \in \prod_{n \in \mathbb{N}} X_n$. We will show that either $X = \emptyset$ or $X = \prod_{n \in \mathbb{N}} X_n$. For each $X_i, i \in \mathbb{N}$, the only open sets are either $\emptyset_i$ or $X_i$. Note that $\prod_{n \in \mathbb{N}} \emptyset_n = \emptyset$, which is again open, by definition of a topology. Moreover, since all but finitely many of the open sets are the whole space, then $\prod_{n \in \mathbb{N}} X_n$ is again open and we have that $X = \prod_{n \in \mathbb{N}} X_n$. Therefore, the only open sets are either $\emptyset$ or $\prod_{n \in \mathbb{N}} X_n$, and we conclude that the product topology on $\prod_{n \in \mathbb{N}} X_n$ is the indiscrete topology.
Fact: If we put the indiscrete topology on $X_n$ any map with codomain $X_n$ is continuous. It follows that any topology on $\prod_n X_n$ will make all projections continuous, and as the product topology is defined as the minimal topology that makes all projections continuous, in that case the product topology on $\prod_n X_n$ is the minimal topology on that set, i.e. the indiscrete topology. Done.
Or, by another standard fact the product topology is generated by the subbase
$$\{\pi_n^{-1}[O]: n \in \Bbb N; O \subseteq X_n \text{ open } \}$$
but all $O$ are $\emptyset$ or $X_n$, so all members of the subbase are also $\emptyset$ and $\prod_n X_n$ and this subbase generates the indiscrete topology.
Note BTW that the index set $\Bbb N$ is irrelevant: it could have been two sets, or $X_i, i \in I$ for any index set $I$. The arguments I gave stay the same.