Why are box topology and product topology different on infinite products of topological spaces?

Let $X_n$ be topological spaces for each $n\in\mathbb{N}$. To avoid the issues pointed out by Najib, assume for each $n$ that $X_n$ is not a point, and the topology on $X_n$ is not the trivial topology (i.e. there is an open set besides $\emptyset$ and $X_n$). For each $n$, let $U_n \subset X_n$ be a proper, nonempty open subset. Then the set $U := \prod\limits_{n\in\mathbb{N}} U_n$ is open in the box topology on $\prod\limits_{n\in\mathbb{N}} X_n$ but not the product topology.

The product topology is generated by sets of the form $\prod\limits_{n\in\mathbb{N}} U_n$ where each $U_n$ is open in $X_n$ and, for all but finitely many $n$, we have $U_n = X_n$. In other words, almost all of the factors have to be the entire space. For the box topology, each factor $U_n$ just has to be open in $X_n$.

Here is one way of understanding why the product topology is more important (even though the box topology seems more intuitive at first). The product topology is the smallest topology such that for each $k\in\mathbb{N}$, the projection map $\pi_k:\prod\limits_{n\in\mathbb{N}} X_n\to X_k$ is continuous. The preimage of an open set $U_k\subseteq X_k$ via $\pi_k$ is one of the basic sets for the product topology described above: specifically, it is $U_k$ in the $k$th factor and the whole space $X_n$ in each other factor. To generate a topology, we need to include finite intersections of such sets (so not the entire space in finitely many positions), but not infinite intersections. So thinking about wanting the $\pi_k$ to be continuous, the product topology has "enough" open sets, and the box topology adds in open sets that aren't needed.

The other two answers [edit: at the time of writing] are essentially correct, but here's a concrete example as to why the box topology has "too many" open sets.

Consider the function $f : \mathbb R \to \mathbb R^\mathbb N$ given by $f(x) = (x,x,x,\dots)$. Consider the subset of $\mathbb R^\mathbb N$ given by $\prod_{n\ge 1}(-2^{-n},\,2^{-n})$. In the box topology this is open. Its pre-image under $f$ is $\{ x \in \mathbb R : \forall n. x \in (-2^{-n},\,2^{-n}) \}$, but it's easy to see that's just $\{ 0 \}$, which is not open. So $f$ is not continuous!

You may or may not decide that this is a surprising result, but the essential property of the product topology is that you can identify continuous functions purely by looking at each individual projection, and with $f$ that is simply not true: every projection is continuous, but $f$ itself is not.

The simplest way, perhaps, to see a particular difference is the product of a finite discrete space.

Consider the product of $\Bbb N$ copies of $\{0,1\}$ (with the discrete topology). The result is the Cantor space. This is a compact metric space, and therefore it has a countable basis, and only $2^{\aleph_0}$ open sets.

Consider the box topology on the same product, then you get a discrete space of size $2^{\aleph_0}$, which therefore has $2^{2^{\aleph_0}}$ open sets, and is most certainly not compact.