Suppose
$$G={\rm Sym}_{{\rm fin}}(\Bbb{N}):=\{\sigma\in S_\Bbb N\mid \sigma(x)=x \text{ for all but finitely many } x\in \Bbb{N}\}.$$
Prove $G$ is not finitely generated.
My solution:
Suppose $G$ is finitely generated, hence $G=\langle S\rangle$ such that $S\subset G$.
Since $S$ is a finite subset and there are $\infty$ numbers in $\mathbb{N}$, there is $x_1\in \mathbb{N}$ such that $\sigma(x_1)=x_1 \forall \sigma\in S$, $\sigma$ is a permutation of $\mathbb{N} .$
Obviously $\forall \sigma\in \langle S\rangle, \sigma(x_1)=x_1.$
On other side, denote $x_2\in \mathbb{N}$.
$\sigma'$ is a permutation of $\mathbb{N}$ $$\sigma'= \begin{cases} x_2,& x=x_1;\\ x_1,& x=x_2;\\ x,& x\neq x_1,x_2;\\ \end{cases}$$
Obviously, $\sigma'\in G$, contradiction.
Hence, $G$ is not finitely generated.
Is my solution correct?
Your solution is correct, but here are just some comments on your proof:
(1) The symbol $\infty$ stands for infinity. So for your second paragraph, you should write "there are infinitely many elements in $\mathbb{N}$", or, $|\mathbb{N}|=\infty$.
(2) On the fourth paragraph, you denote $x_2\in \mathbb{N}$. You should emphasize that $x_2\neq x_1$ so that you can construct a permutation $\sigma'$ later on in order to get a contradiction.
(3) The permutation $\sigma'$ should be written as $$\sigma'(x)= \begin{cases} x_2,& x=x_1;\\ x_1,& x=x_2;\\ x,& x\neq x_1,x_2;\\ \end{cases}$$
(4) Lastly, you may explain why you have the contradiction, which is because $\sigma'\in G$ implies that $\sigma'\in \langle S\rangle$ but $\sigma'(x_1)\neq x_1$, which is absurd due to your observation in third paragraph.