Let $\mathbb{R}_l$ denote the set of real numbers with the lower limit topology having as a basis the collection of all the closed-open intervals of the form $[a, b)$, where $a$ and $b$ are any real numbers such that $a < b$.
Consider $\mathbb{R}_l^2$ with the product topology having as a basis all the Cartesian products of the form $[a, b) \times [c, d)$, where $a$, $b$, $c$, and $d$ are any real numbers such that $a < b$ and $c < d$.
Let $L$ be the subset of $\mathbb{R}_l^2$ given by $$ L \colon= \{\, x \times (-x) \, \colon \, x \in \mathbb{R} \, \}. \tag{Definition 0} $$
How to show that this set $L$ is closed in the product space $\mathbb{R}_l^2$?
Of course, $L$ is a union of (uncountably many) closed sets.
My Attempt:
Let $a \times b$ be any point of $\mathbb{R}_l^2 \setminus L$. Then $a, b \in \mathbb{R}$ such that $b \neq -a$. Refer to (Definition 0) above. Thus either $b < -a$ or $b> -a$.
Case 1. Suppose that $b < -a$. Let $p$ be any real number such that $$ b < p < -a. \tag{1} $$ Then we of course have $$ a < -p < -b. \tag{2} $$
Now consider the basis set $[a, -p) \times [ b, p)$ for the topology of $\mathbb{R}_l^2$ containing the point $a \times b$.
If $x \times y \in [a, -p) \times [b, p)$, then we find that $$ a \leq x < -p, $$ and $$ b \leq y < p. $$ the first of these two relations implies $$ p < -x, $$ and hence we obtain $$ y < p < -x, $$ which implies that $y \neq -x$, and so $x \times y \in \mathbb{R}_l^2 \setminus L$. Therefore we have $$ [a, -p) \times [b, p) \subset \mathbb{R}_l^2 \setminus L. $$
Case 2. Now suppose that $b > -a$. Then $-b < a$. Let $p$ be any real number such that $$ -b < p < a. \tag{3} $$ Then we also have $$ -a < -p < b. \tag{4} $$
Now consider the basis set $[a, a+1) \times [b, b+1)$ for the product topology on $\mathbb{R}_l^2$.
Let $x \times y$ be any point of $[a, a+1) \times [b, b+1)$. Then we have $$ a \leq x < a+1, \tag{5} $$ and $$ b \leq y < b+1. \tag{6} $$
From (3) and (5) we obtain $$ p < x, $$ and hence $$ -x < -p. \tag{7} $$ Now from (4) and (6) we obtain $$ -p < y, $$ which together with (7) implies $$ -x < -p < y, $$ and hence $y \neq -x$, which shows that $x \times y \in \mathbb{R}_l^2 \setminus L$. Therefore $$ [a, a+1) \times [b, b+1) \subset \mathbb{R}_l^2 \setminus L. $$
Thus in either case we find that, for any point $a \times b$ of the set $\mathbb{R}_l^2 \setminus L$, there exists a basis set $B_{a \times b}$ for the product topology on $\mathbb{R}_l^2$ such that $$ a \times b \in B_{a \times b} \subset \mathbb{R}_l^2 \setminus L. $$
Thus $\mathbb{R}_l^2 \setminus L$ is an open set in $\mathbb{R}_l^2$. Hence $L$ is closed in $\mathbb{R}_l^2$.
Is this proof correct? If so, is my presentation correct and clear enough too? Or, are there any issues?
Can we somehow reduce Case 2 in my proof to a without any loss of generality argument following Case 1?