Consider the ordered square $I^2,$ then the set $[0,1]\times [0,1]$ with the dictionary order. Let the general element of $I^2$ be denoted by $x\times y,$ where $x,y\in[0,1].$ The closure of the subset $$S=\{x\times\frac{3}{4}:0<a<x<b<1\}$$ in $I^2$ is
$(A)S\bigcup ((a,b]\times \{0\})\bigcup ([a,b)\times \{1\})$
$(B)S\bigcup ([a,b)\times \{0\})\bigcup ((a,b]\times \{1\})$
$(C)S\bigcup ((a,b)\times \{0\})\bigcup ((a,b)\times \{1\})$
$(D)S\bigcup ((a,b]\times \{0\})$
**My attempt:-
Case 1:-
$a\times b \notin S$. There is an open set containing $a \times b$ which does not intersect $S$(shown in the figure).
Case 2:-
$a\times b \in S$. There is an open set containing $a \times b$ which does not intersect $S\setminus \{a\times b\}$
So, none of the points is a limit point in $I^2$. But Options are different. Can you help me to visualize?
Consider the case you drew, namely: $S= (0,1) \times \{\frac34\}$:
Of course points of $S$ are always in the closure of $S$, but the points with second coordinates $0$ and $1$ require more thought..
Consider what a (basic) neighbourhood of $a \times 0$ where $a >0$ looks like: it's an open interval $I$ with left endpoint $c \times d \in [0,1]\times [0,1]$ and right end point $e \times f$. (only $(0,0)$ and $(1,1)$ have half-open half-closed basic neighbourhoods, as they are the minimum and maximum in the order).
The only way (in the lexicographic order) that $c \times d < (a \times 0)$ can hold, is when $c < a$ ($c=a$ is ruled out, as we cannot go below $0$ in the second coordinate). But then if we pick $c'$ with $c < c' < a$ (wich we can) then $c'\times \frac34$ lies above $c \times d$ and below $a \times 0$ and so lies in $I$. So $S$ intersects all basic open neighbourhoods of $a \times 0$, $a \times 0$ is a limit point of $S$.
If the second coordiante of a point $x \times y$ lies in $(0,1)$ and the point lies outside $S$, we can pick a small open interval around it that misses $S$.
For points of the form $a \times 1$ we can hold a simialr reasoning that all their basic open intervals also intersects $S$. The only points it fails are the said min and max. We do add $0 \times 1$ and $1 \times 0$ as limit points.
The general case that's asked for then has answer $A$, I think.