Is the following Proof Correct?
Proposition. Prove that $\mathbf{Q}$ is a $F_\sigma$ set of $\mathbf{R}$ i.e. it is a union of countably many closed sets in $\mathbf{R}$.
Proof. Consider the subsets $\mathbf{Q}_1,\mathbf{Q}_2,\dots,\mathbf{Q}_n,\dots$ of $\mathbf{Q}$ defined as follows \begin{align*} &\mathbf{Q}_1 = \left\{0,\frac{1}{1},\frac{-1}{1},\frac{2}{1},\frac{-2}{1},\frac{3}{1},\frac{-3}{1},\frac{4}{1},\frac{-4}{1}\dots\right\}\\ &\mathbf{Q}_2 =\left\{0,\frac{1}{2},\frac{-1}{2},\frac{2}{2},\frac{-2}{2},\frac{3}{2},\frac{-3}{2}\dots\right\}\\ &\mathbf{Q}_3 =\left\{0,\frac{1}{3},\frac{-1}{3},\frac{2}{3},\frac{-2}{3},\dots\right\}\\ &\mathbf{Q}_4 =\left\{0,\frac{1}{4},\frac{-1}{4}\cdots\right\}\\ &\text{and so on} \end{align*} A little inspection reveals that by travesing diagonally beginning from the $0$ in $\mathbf{Q}_1$ we may list every conceivable rational number consequently $\mathbf{Q} = \bigcup_{n=1}^{\infty}\mathbf{Q}_n$. To prove the claim in question then it is sufficient to prove that $\mathbf{Q}_n$ is closed for all $n\in\mathbf{N}$, which is immediately apparent when we consider that $$\mathbf{R}\backslash\mathbf{Q}_n = \bigcup_{k=1}^{\infty}\left(-\frac{k+1}{n},\frac{k}{n}\right)\cup\left(-\frac{1}{n},0\right)\cup\left(0,\frac{1}{n}\right)\cup\bigcup_{k=1}^{\infty}\left(\frac{k}{n},\frac{k+1}{n}\right)$$ and is therefore a union of countably many open sets in $\mathbf{R}$.
$\blacksquare$
Yes, your proof is correct.
Let me suggest a simpler proof: $\mathbb{Q} = \bigcup_{q\in \mathbb{Q}} \{q\}$. This is a countable union of closed sets, since $\mathbb{Q}$ is countable, and any singleton is closed.