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.

Your primary question "Is my proof correct?" has been answered by
user $7\cdot 9011\cdot 2\,$.
The ensuing comment-discussion hints at some detour in your proof.
You may directly aim at a bijection by observing that an element in $S$ is either

*

*parallel to the $y$-axis, hence a vertical line having a rational $\,x_0\,$ as $x$-intercept,

*or both its slope $\,m\,$ and its $y$-intercept $\,y_0\,$ are rational.

Taken together, these correspondences define a bijection
$$S\:\longrightarrow\:\big(\{\perp\}\times\mathbb Q\big)\cup\big(\mathbb Q\times\mathbb Q\big)
\:=\: \big(\{\perp\}\cup\mathbb Q\big)\times\mathbb Q\,$$
whose range set is countably infinite.
With the symbol $\perp\,$ standing for "vertical line", and the variable $\,m\in\mathbb Q\,$ as slope we have
$$\big\{(x,y)\in\mathbb R^2 \mid x=x_0\big\}\;\longmapsto\;(\perp, x_0)\\[3ex]
\big\{(x,y)\in\mathbb R^2 \mid y=mx+y_0\big\}\;\longmapsto\;(m, y_0)\,.$$
