They mean all $\langle x,-x+\epsilon\rangle$ such that $x\in(a,b)$ and $0<\epsilon<1/n$; that includes both rational and irrational $x$.
You have a particular $n$ and a non-empty open interval $(a,b)$ such that $(a,b)\subseteq\operatorname{cl}K_n$, where for each $x\in K_n$ we know that $$\left[x,x+\frac1n\right)\times\left[-x,-x+\frac1n\right)\subseteq V\;.$$
Suppose that $x\in(a,b)$ and $0<\epsilon<1/n$. In order to show that $\langle x,-x+\epsilon\rangle\in V$, it suffices to show that there is a $y\in K_n$ such that
$$\langle x,-x+\epsilon\rangle\in\left[y,y+\frac1n\right)\times\left[-y,-y+\frac1n\right)\;.\tag{1}$$
Now $(1)$ holds iff $y\le x<y+\frac1n$ and $-y\le-x+\epsilon<-y+\frac1n$. These can be boiled down to the requirements that $y>x-\frac1n$, $y\ge x-\epsilon$, $y\le x$, and $y<x-\epsilon+\frac1n$. Can you continue from here to show that such a $y$ can always be found? I’ve finished the argument in the spoiler-protected bit below.
Since $\epsilon<\frac1n$, these four inequalities further reduce to $y\ge x-\epsilon$ and $y\le x$, i.e., $x-\epsilon\le y\le x$. Since $\epsilon>0$, the open interval $(x-\epsilon,x)$ is non-empty, and it follows that $(x-\epsilon,x)\cap(a,b)\ne\varnothing$. Finally, $K_n$ is dense in $(a,b)$, and $(x-\epsilon,x)\cap(a,b)$ is a non-empty open subset of $(a,b)$, so $(x-\epsilon,x)\cap(a,b)\cap K_n\ne\varnothing$. Now just choose $y$ in this intersection, and you’re done.
It’s not true that all points of $\operatorname{cl}K_n$ lie in $V$, but the points of $K_n$ are so thickly scattered throughout the interval $(a,b)$ that
$$\{x\}\times\left(-x,-x+\frac1n\right)\subseteq V$$
for each $x\in(a,b)$, even when $x$ itself isn’t in $K_n$. For each $y\in\left(-x,-x+\frac1n\right)$ there is a point $z\in K_n$ to the left of $x$ that is close enough to $x$ that $$\langle x,y\rangle\in\left[x,x+\frac1n\right)\times\left[-x,-x+\frac1n\right)\;,$$ which of course is a subset of $V$.
HINT: Prove that the Sorgenfrey line is Lindelöf, and use the fact that a regular Lindelöf space is paracompact. To prove that it’s Lindelöf, start with a basic open cover $\mathscr{U}$ (i.e., a cover by sets of the form $[a,b)$). Show that $\{(a,b):[a,b)\in\mathscr{U}\}$ covers all but a countable subset of $\Bbb R$, and use the fact that $\Bbb R$ with the usual topology, being second countable, is hereditarily Lindelöf.
Best Answer
Normality only demands that we be able to separate arbitrary disjoint closed sets - since $\mathbb{Q}$ is neither closed nor open in the Sorgenfrey line (or in the usual topology on $\mathbb{R}$ for that matter), our inability to separate the rationals and the irrationals by closed neighborhoods is irrelevant.