# Follow-up: that the closures of every two open sets must intersect

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.

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The closure of $N_\epsilon((x,y))$ is the set of all points $p$ such that every neighbourhood of $p$ contains a point of $N_\epsilon((x,y))$. The neighbourhoods of a point $p$ are themselves of the form $N_\delta(p)$ and consist of two intervals on the rational $x$ axis. If $p$ lies within one of the four strips, its projection along one of the two directions lies in an interval $I$ of the intervals forming $N_\epsilon((x,y))$. If so, the interval of the two intervals forming $N_\delta(p)$ that is centred around this projection overlaps with $I$ for any $\delta$. Thus such $p$ is in the closure of $N_\epsilon((x,y))$. Conversely, if $p$ does not lie in one of the four strips, its projection along both directions lies outside both of the intervals forming $N_\epsilon((x,y))$, so we can choose $\delta$ small enough that $N_\delta(p)$ does not overlap with $N_\epsilon((x,y))$.