A better way to think about $\mathbb{R^N}$ is as the set of all real-valued sequences, right? You have $\mathbb N$ dimensions and each dimension corresponds to a real number, so $\mathbb{R^N}$ is just the set of all real-valued sequences (as a note $\mathbb{R^R}$ can be thought of as all the functions from $\mathbb R$ to $\mathbb R$).
So with this definition in mind, what does it mean for there only be a finite number of non-zero indexes in a sequence. Well it means that past a certain point the sequence is entirely zeros. Okay now keeping this in mind, think about what the product topology means. Well it means that any open set (We'll just work with basis sets) is a product of spaces $X_i$ where for only finitely many $i$ $X_i\neq \mathbb R$. Then if we want to show that our set is dense in $\mathbb{R^N}$, we need to show that it intersects every basis open set. Well we've just said that an open set has only finite many indices that aren't the whole space. So if we pick an $x \in X_i$ where each $X_i \neq \mathbb R$, then we'll get finitely many points that aren't zero and we can let everything else be zero. This will show that our set is dense in $\mathbb{R^N}$.
What happens if we put the box topology on $\mathbb{R^N}$?