I was reading through general-topology posts, but I couldn't understand the reasoning behind the answer of this part of a post. I'll restate it here:
This concerns the irrational slope topology. That is, let $X = \{(x,y)\in\mathbb{Q}^2:y\geq 0\}$. Fix $\theta\in\mathbb{R}\backslash\mathbb{Q}$. Let $\mathcal{T}$ be the coarsest topology on $X$ containing the sets of the form $$N_\epsilon(x,y)=\{(x,y)\}\cup\{(q,0)\mid q\in\mathbb{Q},\left|q-\left(x+\frac{y}{\theta}\right)\right|<\epsilon\}\cup\{(q,0)\mid q\in\mathbb{Q},\left|q-\left(x-\frac{y}{\theta}\right)\right|<\epsilon\}.$$ for $(x,y)\in X$ and $\epsilon>0$. (The following $B_\epsilon$'s correspond to the latter two sets in the logical manner.)
Claim 1. The closure of each basis neighborhood $N_\epsilon((x,y))$ contains the union of the four strips of slope $\pm\theta$ emanating from $B_\epsilon(x+y/\theta)$ and $B_\epsilon(x−y/\theta)$.
Knowing this, why does it then follow that Claim 2 holds?
Claim 2. The closures of any two open sets intersect (nontrivially).
I think my problems stems largely from the fact that I don't understand the boxed claim to begin with. My interpretation of the claim is that each basis neighborhood must contain all 4 line segments, but this doesn't make much sense:
The basis neighborhoods are merely (open) intervals on the $x$-axis unioned with the point $(x,y)$, whereas each line segment is a line in $\mathbb{R}^2$ with irrational slope (starting at $(x,y)$ and ending at the $x$-intercept). Surely, I'm making a mistake? I appreciate any help.