${\mathbb{R}}^{\mathbb{N}}$ in the box topology is disconnected, the bounded and the unbounded sequences forming a separation, although every factor is connected and the product topology is connected

Refuted: that a product of connected spaces is connected in the box topology. A product of connected spaces is connected in the product topology, and that argument is a theorem of ZF; for an infinite index set it is the assertion that the product of nonempty spaces is nonempty that uses the Axiom of Choice proves this for the product topology only, and the restriction is not a matter of convenience.

Witness. Let ${\mathbb{R}}^{\mathbb{N}}:={\prod }_{n\in \mathbb{N}}\mathbb{R}$, each factor carrying the usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and give it the box topology ${\mathcal{T}}^{\mathrm{\square }}$ (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). A point of ${\mathbb{R}}^{\mathbb{N}}$ is a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Put

$B:=\{x\in {\mathbb{R}}^{\mathbb{N}}:x\text{ is bounded}\},V:=\{x\in {\mathbb{R}}^{\mathbb{N}}:x\text{ is unbounded}\}$

(Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set). Then $(B,V)$ is a separation of ${\mathbb{R}}^{\mathbb{N}}$ in the box topology (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets), while every factor $\mathbb{R}$ is connected and ${\mathbb{R}}^{\mathbb{N}}$ is connected in the product topology (A product of connected spaces is connected in the product topology, and that argument is a theorem of ZF; for an infinite index set it is the assertion that the product of nonempty spaces is nonempty that uses the Axiom of Choice).

• The proof of the product-topology theorem breaks at exactly one place. There the finite-support points around a base point are dense, because a basic product-open set constrains only finitely many coordinates. A box constrains every coordinate at once, so a point differing from the base point in finitely many coordinates need not lie in a given box, the density argument fails, and with it the conclusion.

• The separating property is invariant under bounded perturbation, and that is all that is needed. Boundedness of a sequence is unchanged by moving every coordinate by less than $1$, and a single box of width $2$ around a point performs exactly such a perturbation. Any property with that stability separates the box topology in the same way.

• Nothing here contradicts The box topology is finer than the product topology, the two agree for a finite index set in ZF, and, assuming the Axiom of Choice for nonempty factors, the box topology is strictly finer whenever infinitely many factors have a nonempty proper open subset. The box topology is finer, so it has more open sets and therefore more chances to separate; a finer topology can disconnect a space that a coarser one connects, and this witness is that phenomenon in its simplest form.