This is a standard trick. The proof is as follows:
Let [tex]A_0=\mathbb{Q}[/tex]. This is clearly countable.
Let [tex]A_1=\{a+bx~\vert~a,b\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times \mathbb{Q}[/tex].
Let [tex]A_2=\{a+bx+cx^2~\vert~a,b,c\in \mathbb{Q}\}[/tex], this is countable since it has the same cardinality of [tex]\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}[/tex].
Continue with this process. Finally, we have [tex]\mathbb{Q}[x]=\bigcup_{n\in \mathbb{N}}{A_n}[/tex]. This is countable as countable union of countable sets.