We have set $S=\{ (a,b) | a,b \in \mathbb{Q}\}$
And we know that $(a,b)\sim \mathbb{R}$ , so $k((a,b))=c$. And $\mathbb Q \sim \mathbb N$, so $k(\mathbb Q)=\aleph_0$.
I don't know how to put all informations together.
I was also thinking that $S \subset \mathbb R$ and than we know $k(S) \le k(\mathbb R)=c$. But how to get other inclusion.
On
You're mixing the cardinality of the interval, with the cardinality of the set of intervals.
Note that while $\Bbb R$ has cardinality $2^{\aleph_0}$, $\{\Bbb R\}$ has cardinality $1$. It is a singleton.
So to see that $\{(a,b)\subseteq\Bbb R\mid a,b\in\Bbb Q\}$ is countable, note that each such interval is uniquely identified by its endpoints, which gives us a very natural injection (although not a bijection) into $\Bbb Q^2$.
On
Let us start with a simple example. The set of intervals $\{(0,1),(1,2),(2,3)\}$ contains only three intervals. You can easily define a bijection with the set of three integers: $\{0,1,2\}$, with cardinal $3$: $$(0,1) \leftrightarrow 0$$ $$(1,2) \leftrightarrow 1$$ $$(2,3) \leftrightarrow 2$$ No matter if $(0,1)$ is an interval. Similarly, the set of segments $\{(a,b)| a,b \in \mathbb{Q}\}$ can be put in bijection with the set of ordered pairs $\{[a,b]| a,b \in \mathbb{Q}\}$, whose cardinality is $\mathbb{Q}^2$. Consider that the notation $[a,b]$ stands here for ordered pairs, since you have used $(a,b)$ for intervals
Hint: there exists an injective map $f: S \to \mathbb{Q}^2$. Since $\mathbb{Q}^2$ is countable, $S$ is at most countable.