In order to give an example of a topology of $\Bbb{R}^2$ which is not the product of a topology of $\Bbb{R}$, I have seen on a book the topology $\mathcal{T}$ of semi-open triangles induced by the basis $\mathcal{B} = \{ V(p,\varepsilon) : p \in \Bbb{R}^2, \varepsilon > 0 \}$ where given $p=(a,b)$ and $\varepsilon > 0 $ we denote $$V(p,\varepsilon) := \{(x,y) \in \Bbb{R}^2 : a \leq x < a + \varepsilon \, ; \, b - (x - a) \leq y \leq b + x - a \}$$
The book affirms that this topology can not be described as a product topology but it does not show it, so if anyone knows a proof or another source to consult, I would appreciate it so much.
Hint: Prove that the subspace topology on every vertical line is discrete and the subspace topology on every horizontal line is the lower limit topology (LLT). Also prove that your topology is not the product of discrete topology and the LTT. Then finish the proof that this topology is not a product topology.