Is $[ \sqrt 2, \sqrt 3] \cap \mathbb{Q}$ an open subset of $\mathbb{Q}$?
No, that's wrong. The fact that a set is closed doesn't mean it is not open!
In fact $K$ is also open because it equals to $(\sqrt{2},\sqrt{3})\cap \mathbb{Q}$.
Side note: The space $\mathbb{Q}$ with the topology induced by $\mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.
Yes, $K$ is an open subset of $\mathbb Q$, since $K=\left(\sqrt2,\sqrt3\right)\cap\mathbb Q$ and $\left(\sqrt2,\sqrt3\right)$ is an open subset of $\mathbb R$.
$K=[\sqrt 2,\sqrt 3]∩\mathbb{Q}=\{q\in \mathbb{Q}|\sqrt 2<q<\sqrt 3\}$ where$[\sqrt 2,\sqrt 3]$ is closed in R.
From this I can conclude that K is not open subset of Q
You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:
K is an intersection between a closed set and a closed set.
K is therefore closed.
Therefore K is not open.
The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.
If we have the open ball topology, then since $\sqrt2 <q$, we know that there is "space" between $\sqrt 2$ and $q$, and similarly for $\sqrt3$. So given any $q$, we can take $\epsilon_1$ to be half the distance between $\sqrt2$ and $q$, $\epsilon_2$ to be half the distance between $\sqrt3$ and $q$, and $\epsilon$ to be the minimum of $\epsilon_1$ and $\epsilon_2$. Then everything withing $\epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.