Open sets on ordered topology in $\mathbb{R\times R}$

The set $U:=(0,1)\times(0,1]$ is not open.

Proof.

Any neighborhood of the point $\langle 1/2, 1\rangle\in U$ contains some point $\langle 1/2,y\rangle$ with $y>1$ which is therefore not in $U$. So not all points of $U$ are inner points and $U$ cannot be open.

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You can think about the basic open sets $\{(x,y)\in\Bbb R^2\mid (x_1,y_1)<(x,y)<(x_2,y_2)\}$ of your space as some uncountable generalization of the following constellation:

The colored region should represent the "open" set. Note that the first coordinate is the top-down-direction, the second one the left-right-direction (might be flipped compared to usual mathematical plots). These open sets have (at most) two incompletely filled "lines", the upper and the lower one. All "lines" in between are completely filled. The first and the last line have open ends, something which is not representable in this discrete visualization.

In this sense, you see that open sets (that span several lines) include "whole lines". The basic open sets (that span several lines) can be decomposed as follows:

\begin{align} \{(x,y)\in\Bbb R^2\mid &(x_1,y_1)<(x,y)<(x_2,y_2)\} \\[0.5em]&= \color{red}{\underbrace{\{x_1\} \times (y_1,\infty)}_{\text{upper half line}} } \;\;\cup \color{blue}{\underbrace{(x_1,x_2)\times \Bbb R}_{\text{whole lines in between}} } \cup\;\;\color{green}{\underbrace{ \{x_2\}\times (-\infty, y_2)}_{\text{bottom half line}}} \end{align}

as highlighted in the following discrete visualization:

The topology that you have described is called the dictionary order topology on $\mathbb R \times\mathbb R$.

Assume that you are a librarian, any you have two books in your hand with inner library code $C3$ and $E2$.

Now, what first you do is that you compare the alphabets of the library codes of the books, namely you determine whether $C <, >, = E$ in the natural sense. In this case, since $C < E$, you do know that in a shelf starting from left and going to right, you put the first book, with the code $C3$, to the left of the second book, with the code $E2$, in that shelf.

Now, consider the books $D4$ and $D2$.To determine the order of the book in the shelf, if you compare the alphabets, they are equal, so what you, you check the numbers, since $4 > 2$, the first book, with the code $D4$, should be in the right of the second book, with the code $D2$, in the shelf.

Now, coming to you question, in the dictionary order topology $\mathbb R \times\mathbb R$, you do the same thing, given two elements $(a,b)$ and $(c,d)$, you first compare the first inputs, or say components, of the elements, namely you check whether $a <, >, = c$ in the usual sense, and if $a = c$, you check the second component of the elements, and determine which element is "less" than other element.

Now, in this way, you order all the elements of $\mathbb R \times\mathbb R$, and in an order topology, the open sets of that topology are the intervals, i.e (e,f), where $e,f \in \mathbb R \times\mathbb R$.

In particular, in your example, the set $$(0, 1)\times(0,1] = \{(x,y) \in \mathbb R \times\mathbb R| 0 <x<1, 0<y \leq 1 \} \\ = \cup_{x \in (0,1)} \{(x,y) | 0 < y <1\} \cup_{x \in (0,1)} \{(x,1)\}$$

and since the set contanining only the points of the form $(x,1)$ where $x \in (0,1)$, is not open, its union with some open set is not open.