Cardinality of set of all straight lines in $\mathbb R^2$, that pass through at least two different rational coordinates. Definition: A coordinate $(x,y)\in \mathbb R^2$ is rational if $x,y$ are both rational nos. 
Let $S$ be the set of all straight lines in $\mathbb R^2$ that pass through at least two distinct rational points. 
Let $C$ be the set of all rational points in $\mathbb R^2$. Since $\mathbb Q \times \mathbb Q$ is countable so are its subsets. Therefore, it is possible to write  $C=\{c_1,c_2,...,c_k,c_{k+1},...\}$. Of course $C=\mathbb Q \times \mathbb 
 Q$.
Let $S_i$ be the set of all straight lines that pass through $c_i\in C$. For every $s\in S_i$, $\exists$ a rational point $c_{k(s)}\ne c_i$ thus we can have a bijection. Therefore, $S_i$ is countable.
Hence, $S=\cup_{i=1}^{\infty} S_i$ is countable collection of countable sets and therefore countable. 
Is my proof correct? Thanks.

Yes, but not completely:

*

*The set $S_i$, as defined, is not countable. In order to be countable, the set $S_i$ should be defined as the set of lines in $S$, tha passes through $c_i$.


*also, with this definition, $S_i$ the described map is not a bijection. It is true that there is a map from $C$ to $S_i$ that assigns to $c_j$ the line passing through $c_i$ and $c_j$. But this map is not a bijection (it is not injective): Still, it is surjective. This implies that $S_i$ is countable.
Last step of the proof is now correct.
