Consider the topology on $\mathbb R$ generated by the basis $\{[a, b) : a,b \in \mathbb Q\}$. What is the closure of the set $(0,\sqrt 2)$? 
Consider the topology on $\mathbb R$ generated by the basis $\left \{[a,b) : a,b \in \mathbb Q\right \}$. Denote the topology by $\Gamma_l$. What is the closure of the set $(0,\sqrt 2)$ in $\Gamma_l$?

The guess candidates are $\left [0,\sqrt 2\right )$, $\left (0,\sqrt 2\right ]$ and $\left [0,\sqrt 2\right ]$. Out of these, I could eliminate $\left [0,\sqrt 2\right )$ since
$$\left [0,\sqrt 2\right )=\bigcup_{n=1}^\infty \left [0,q_n\right )$$
where $\{q_n\}_{n=1}^\infty$ is an increasing sequence of rationals converging to $\sqrt 2$, and $q_1>0$.
[Note: As Theo pointed out, it does not eliminate the possiblity. Being open does not imply that it's not closed]
But, I am unable to perform such manipulations on $\left (0,\sqrt 2\right ]$ or $\left [0,\sqrt 2\right ]$. I tried to look at the complement of $\left (0,\sqrt 2\right ]$ which is $(-\infty,0]\cup\left (\sqrt 2,\infty\right )$ one of which is closed and the other is open.
I would like to know whether there are any tricks to guess the closure of such sets in these weird topologies, and a proof for the current problem.
 A: As a hint for the procedure: if $A$ is the set you want to find the closure of, consider all points $p \notin A$. If it is the case that any basic open set containing $p$ must intersect $A$, then $p$ is an adherence point of $A$ and is in the closure (all points of $A$ of course as well, but we don’t have to test them).
As an illustration I''ll do the case $A=(0,\sqrt{2})$.
If $x > \sqrt{2}$ we find some $q \in \Bbb Q$ so that $\sqrt{2} < q < x$ by order denseness of $\Bbb Q$. Also take any rational $q_2 > x$ and note that $x \in [q,q_2)$ and this neighbourhood is disjoint from $A$ so that it is clear that $x \notin \overline{A}$.
For $x=\sqrt{2}$ a basic open neighbout of $x$ is of the form $[q_1, q_2)$ with $q_1 \le \sqrt{2} < q_2$. In fact $q_1 < \sqrt{2}$ and we can find a real $x'>0$ in $(q_1, \sqrt{2})$. This $x$ lies in $A \cap [q_1, q_2)$ so that we have shown that every basic neighbourhood of $\sqrt 2$ intersects $A$, so $\sqrt{2} \in \overline{A}$.
If $x <0$ we find a rational $q \le x$ and then $[q, 0)$ is a basic neighbourhood missing $A$.
If $x=0$ we have that a basic neighbourhood of $0$ can be assumed to be of the form $[0,q)$ with $q >0$ rational. WLOG $q < 2$, say and then $\frac{q}{2} \in [0,q) \cap A$ so also $0 \in \overline{A}$. Hence $\overline{A}=[0,\sqrt 2]$.
