I am asking about this question:
Checking an example of a monotone class.
1-Why in the example of the OP $(-\infty, a) \cup [a, \infty)$ not equal to $\mathbb{R}$?
2-How can I prove rigorously that $\{0\}$ does not belong to $$\{(-\infty, a_{n}), (-\infty, a_{n}], \emptyset, [a_{n}, \infty),\mathbb{R}, (a_{n}, \infty)\}, $$I believe that the answer of Kavi there is very convincing but is not their a rigorous proof?
You have misunderstood that question. The claim made in the question is that $$\mathcal{G} = \{(-\infty, a), (-\infty, a], \emptyset, [a, \infty), (a, \infty): a \in \mathbb{R}\}$$ is itself a monotone class. The observation you make that $\mathbb{R} = (- \infty, a) \cup [a, \infty)$ is exactly the problem that the answer points to. If $\mathcal{G}$ were a monotone class then this would imply that $\mathbb{R} \in \mathcal{G}$. However $\mathbb{R} \not \in \mathcal{G}$ so that $\mathcal{G}$ cannot be a monotone class.
As for the second question you ask, from the definition of $\mathcal{G}$, if $A \in \mathcal{G}$ then there is $a \in \mathbb{R}$ such that $A \in \{(-\infty,a), (-\infty,a], (a, \infty), [a, \infty)\}$. So since $\{0\}$ is non-empty, you just need to check it is not one of these half-intervals for some $a$. This is obvious since all of the half-intervals contain uncountably many elements.