Why is $(-\infty, 1)$ open in $\mathbb{R}_{\ell}$? Let $\mathbb{R}_l$ be the set $\mathbb{R}$ of real numbers associated with the topology given by the basis $\mathcal{B} = \{ [a, b); a < b, a, b \in \mathbb{Q}\}$. Determine the closure of the subsets $(1, \sqrt{2})$ and $(\sqrt{2}, 3)$ in $\mathbb{R}_l$
Here's my attempt:

Closure of $A = (1, \sqrt{2})$ in the topology generated by the basis $\mathcal{B}$ is $[1,\sqrt{2}]$. First of all, this set is closed. To prove this, we just need to show that its complement is an open set. We have $X \setminus[1, \sqrt{2}]= (-\infty, 1) \cup (\sqrt{2}, \infty)$. Note that the arbitrary union of open sets is open, so we just have to show that $(-\infty, 1)$ and $(\sqrt{2}, \infty)$ are open. Now, for any $\varepsilon >0$, every neighborhood $[1, 1+ \varepsilon )$ around $1$ intersect $A$, so $1$ is an element of $A$-closure. There do not exist open sets like $[\sqrt{2}, 4)$, where $\sqrt{2}$ is the lowest element in the set.  Thus, any open set containing $\sqrt{2}$ must contain an element lower than $\sqrt{2}$.  Thus $\sqrt{2}$ is an element of $A$-closure.
Closure of $B = (\sqrt{2}, 3)$ in the topology generated by the basis $\mathcal{B}$ is $[\sqrt{2}, 3)$.  Every neighborhood around $\sqrt{2}$ must intersect $B$, so $\sqrt{2}$ is an element of $B$-closure.  But, for any $\varepsilon >0$, $[3, 3+\varepsilon )$ is an open set containing $3$, thus $3$ is not an element of $B$-closure.

And I'm stuck in the part to show that these sets are closed or the complements are open. For example, is it ok to say that for any $n \geq 0$, $[-\frac{1}{n}, 1)$ is open, so $(-\infty, 1) =\lim_{n\to 0} [-\frac{1}{n},1)$ is open? And also to show that $(\sqrt{2}, \infty)$ is open, can I say that there is $a \in \mathbb{Q}$ such that $a > \sqrt2$ and $[a, \infty)$ is open?

In a topology, it is necessary that a countable union of open sets is again open. So the goal here would be to write $(-\infty, 1)$ as a countable union of open sets $[a,b)$. Formally, we have
$$ (-\infty, 1) = \displaystyle\bigcup_{n=1}^\infty [a_n , b_n),$$
with each $[a_n , b_n)$ an open set in $\mathbb{R}_l.$
Your expression $(-\infty, 1) = \displaystyle\lim_{n\to 0} [-1/n, 1)$ is similar to this expression, but I would suggest adjusting your right hand side to be a limit as $n$ goes to $\infty$ for clarity. You may then see an expression that is more in accordance with the definition of openness in a topology.
"Can I say that there is $a\in\mathbb{Q}$ such that $a > \sqrt{2}$ and $[a, \infty)$ is open?"
This statement is true and may be helpful for showing that $(\sqrt{2}, \infty)$ is open, though you should ask yourself what property of the rational numbers allow you to say this. Given this information, can you express $(\sqrt{2}, \infty)$ as a countable union of open sets? Try to do so, and compare your countable union to the given set $(\sqrt{2}, \infty)$. Note in particular that $(\sqrt{2}, \infty)$ is open on the left, 'open' being in the usual sense.
