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 $V$.
Here's a simple direct proof, which works just as well for the Sorgenfrey topology as for the usual topology of the line.
Let $\mathcal U$ be a collection of Sorgenfrey-open sets that covers $\mathbb R$. Let's say that a set $X\subseteq R$ is countably covered if $X$ is covered by countably many members of $\mathcal U$. We want to show that $\mathbb R$ is countably covered.
Consider any $a\in\mathbb R$, and let $C_a=\{x: x\ge a,\text{ and the interval }[a,x]\text{ is countably covered}\}$. It's easy to see that $\sup C_a=\infty$; assuming the contrary leads to a contradiction. Hence every finite interval $[a,b]$ is countably covered, and so is $\mathbb R=\bigcup_{n\in\mathbb N}[-n,n]$.
P.S. I have been asked to explain why assuming that $\sup C_a=b\in\mathbb R$ leads to a contradiction. Let $b_n=b-\frac{b-a}{2^n}$ for $n=1,2,3,\dots,$ so that $a\lt b_n\lt b$ and $b_n\to b.$ Thus for each $n$ there is a countable collection $\mathcal S_n\subseteq\mathcal U$ such that $[a,b_n]$ is covered by $\mathcal S_n,$ and the half-open interval $[a,b)$ is covered by the countable collection $\bigcup_{n\in\mathbb N}\mathcal S_n.$ Moreover, since $\mathcal U$ covers $\mathbb R,$ there is some $U\in\mathcal U$ such that $b\in U.$ Since $U$ is Sorgenfrey-open, there is some neighborhood $[b,b+\varepsilon)$ of $b$ (with $\varepsilon\gt0$) such that $[b,b+\varepsilon)\subseteq U.$ Then $[a,b+\varepsilon)$ is covered by $\{U\}\cup\bigcup_{n\in\mathbb N}\mathcal S_n,$ whence $b+\frac\varepsilon2\in C_a,$ contradicting our assumption that $b=\sup C_a.$
Best Answer
$Z \times Z$ is indeed normal. To see this, observe that $Z$ is second countable and normal. In particular, $Z$ is regular. Thus, $Z\times Z$ is regular and second-countable. Finally, notice that a regular second-countable space is normal.