Let $$g:\mathbb{R}^2 \longrightarrow \mathbb{R} (x,y) \longmapsto xy$$ be the multiplication map. I want to show $g$ is open -- it maps open sets to open sets.
My approach is to take basis elements of the product topology, which look like $A = (a,b) \times (c,d)$ where $(a,b),(c,d)$ are open intervals in $\mathbb{R}$. It is intuitively obvious that if
$$u = \text{min}(ac,ad,bc,bd)$$ $$v = \text{max}(ac,ad,bc,bd)$$
Then $g(A) = (u,v)$, an open interval. If $U$ is open in $\mathbb{R}^2$, it is a union of these basis elements, so $g(U)$ is the union of open intervals and thus open.
I don't feel like this is substantial enough to be a proof though. Is there a way to do this elegantly without considering many cases? Should I start in a different place? The same problem is here but isn't helpful to me.
For bonus points, $g$ is not closed. Take
$$B = \Big\{ \Big(x , \frac{\text{arctan}(x)}{x} \Big) | x \in \mathbb{R} - \{0\} \Big\}$$
Then $B$ is closed but $g(B) = (-\frac{\pi}{2},\frac{\pi}{2})$, which isn't closed.
Your idea is correct: Show that there exists a basis whose members are mapped by $g$ to open sets. And in fact, the open squares form such a basis. However, you need some arguments to prove that $g((a,b) \times (c,d)) = (u,v)$. You omitted that, but admittedly it not too difficult. Certainly you have to take special care of the cases $0 \in (a,b)$ and $0 \in (c,d)$. Once you have done this, your proof is complete.