When assuming the axiom of choice the product topology and box the topology are quite different when considering infinite products. For example the Tychonoff product of countably many sets of three elements is compact, metrizable an all in all a very nice space. On the other hand, the box product is not separable or second countable at all.

But without the axiom of choice the world is indeed a strange place. This was posted as answer on math.SE earlier today.

**Definition.** We say that $S$ is a Russell set if there exists a partition of $S$ into $P_n$ for $n\in\omega$, such that $|P_n|=2$ for all $n$, and for every infinite $I\subseteq\omega$, the product $\prod_{i\in I}P_i$ is empty.

Russell sets are infinite Dedekind-finite sets. They cannot be linearly ordered, and therefore witness both the failure of the axiom of countable choice (even for families of pairs, which is the least you can ask!) and the Boolean Prime Ideal theorem (which implies every set can be linearly ordered).

The existence of Russell sets is consistent, for example in Cohen’s second model there are Russell sets. Work in such model where a Russell set exists. Let $S$ be a Russell set, and $\{P_n\mid n\in\omega\}$ its partition into pairs witnessing this. We may assume that $S\cap\omega=\varnothing$, now define $X_n=P_n\cup\{n\}$ and $T_n=\{\varnothing,P_n,X_n\}$ is a topology on $X_n$.

Note that $\prod X_n$ is non-empty, but every choice function must pick co-finitely many ordinals. Therefore the space $\prod X_n$ is non-empty and so we can meaningfully define a topology on that space. Now look at the box topology defined there. If $U_n$ is a sequence of open sets, then $\prod U_n$ is non-empty if and only if $U_n\neq\varnothing$ for all $n$, and cofinitely often $U_n=X_n$.

This means that the box product is exactly the Tychonoff product!

Weird, huh?