Let $R=\mathbb{Q}[x_1,x_2,\dots]$ and $M=\mathbb{Q}$. Show that $M$ is finitely generated $R$-module.
First of all, I don't understand how is $M$ an $R$-module; Indeed $f(x_1,x_2,\dots)=x_1\in R, 1\in M$ but $1\cdot f(x_1,x_2,\dots)=x_1\notin \mathbb{Q}=M$.
Is this a mistake in the question.
Assuming it's a mistake we need to change $R\leftrightarrow M$. I still don't know how to solve it because we need to find a finite basis $\{m_1,\dots ,m_n\}\subseteq M$ s.t. $M=\{\sum_{i=1}^nr_im_i:r_i\in R\}$.
The netural guess is to detemine $m_i:=x_i$ but this way $x_{n+1}\notin \operatorname{span} \{x_1,\dots,x_n\}$.
An $R$-module $M$ is an abelian group equipped with a ring homomorphism $$\varphi:\ R\ \longrightarrow\ \operatorname{End}(M).$$ In this particular case, a ring homomorphism $$\varphi:\ \Bbb{Q}[x_1,x_2,\ldots]\ \longrightarrow\ \operatorname{End}(\Bbb{Q}),$$ is uniquely determined by where the indeterminates $x_i$ are mapped, and an endomorphism of $\Bbb{Q}$ (as an abelian group!) is uniquely determined by where $1\in\Bbb{Q}$ is mapped. So $\varphi$ is uniquely determined by the values $(\varphi(x_i))(1)\in\Bbb{Q}$.
Note that here $\varphi(x_i)$ is an endomorphism of $\Bbb{Q}$ for each $i$. In your notation $(\varphi(x_i))(1)=1\cdot x_i$. Recall that the '$\cdot$' in $1\cdot f(x_i)$ stands for the 'multiplication' defined by the $R$-module structure on $M$; it is not the regular multiplication $1\cdot f(x_i)\in\Bbb{Q}[x_1,x_2,\ldots]$ because we are not in $\Bbb{Q}[x_1,x_2,\ldots]$, we are in $\Bbb{Q}$ with a $\Bbb{Q}[x_1,x_2,\ldots]$-module structure.