The function $f:\mathbb R \to \mathbb R^\infty$ given by $f(t)=(t,t,t,t,t\dots)$ is said to be not continuous in the box topology but continuous in the product topology.
The example given is $U=(-1,1)\times(-1/2,1/2)\times (-1/3,1/3)\times \dots $. Why does this set show the box topology is not continuous by $f$? We can just take the preimage of $U$ and it would just be $U$ correct? I can see that $(-1/n,1/n)$ converges to $0$ which is closed in $\mathbb R$, but for any specific $\mathbb R_i$ in $\mathbb R^\infty$, we actually do have an open interval even if it is extremely small.
The preimage of $U$ is a set in $\mathbb{R}$ and since $f$ is the identity function in each coordinate, it is given by all the points in $\mathbb{R}$ such that they are smaller than $\frac{1}{n}$ for every $n$. Hence you easily see that the preimage is the singleton {0}, which is not open since if it were open, it would contain some interval $(- \delta , \delta)$ for $\delta > 0$. But this would mean that $f( (- \delta , \delta) )\subset U$, getting for each coordinate $(- \delta , \delta) \subset (- 1/n , 1/n)$ for every $n$, reaching a contradiction (since if it holds for every $n$, then it holds also in the limit that here is zero).
Note that the difference with the product topology is that with the product topology we have only finitely many intervals of the type $(- 1/n , 1/n)$, hence we will always a be able to find a small $\delta > 0$ such that $(- \delta , \delta) \subset (- 1/n , 1/n)$ since now $n$ can take only finitely many values, and you cannot take the limit as $n$ going to infinity.