Let =[0,1]×[0,1] with the lexicographic order topology and ={(1/)×0|∈ℤ+}
I already read this Finding the closure of some subsets of the ordered square
what is open neighborhood of <0, 1>??
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The point $\langle0,1\rangle$ doesn't belong to $B=(0,1]\times[1/2,1]$, because the first component of the pairs in $B$ is different from zero (since the first coordinate takes value in $(0,1]$, that is the unit interval minus zero) – Alessandro Apr 17 '21 at 08:01
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A basic neighbourhood of $(0,1)$ is an open interval $\langle (a,b), (c,d) \rangle$ that $(0,1)$ lies in so $(a,b)$ is of the form $(0,t)$ for some $t < 1$ (so that it is smaller than $(0,1)$) and $(c,d)$ has $c > 0$ and $d \in [0,1]$ (to be larger than $(0,1)$, but then some $0 < \frac1n < c$ exists and $(\frac1n,0) \in \langle (a,b), (c,d) \rangle$ for that $n$, and so every basic neighbourhood of $(0,1)$ intersects $A$. So $(0,1) \in A'$ as claimed.
Your set $[0,1] \times [\frac12,1]$ is not a neighbourhood of $(0,1)$.. The basic interval $\langle (a,b), (c,d) \rangle$ is of the form $(\{0\} \times (t,1]) \cup ((0,c] \times [0,d))$, e.g.
Henno Brandsma
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