Proof That $\mathbb{R} \setminus \mathbb{Q}$ Is Not an $F_{\sigma}$ Set

I am trying to prove that the set of irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ is not an $F_{\sigma}$ set. Here's my attempt:

Assume that indeed $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then we may write it as a countable union of closed subsets $C_i$: $$ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i $$ But $\text{int} ( \mathbb{R} \setminus \mathbb{Q}) = \emptyset$, so in fact each $C_i$ has empty interior as well. But then each $C_i$ is nowhere dense, hence $ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i$ is thin. But we know $\mathbb{R} \setminus \mathbb{Q}$ is thick, a contradiction.

This seems a bit too simple. I looked this up online, and although I haven't found the solution anywhere, many times there is a hint: Use Baire's Theorem. Have I skipped an important step I should explain further or is Baire's Theorem used implicitly in my proof? Or is my proof wrong? Thanks.

EDIT: Thin and thick might not be the most standard terms so:

Thin = meager = 1st Baire category

Your solution is correct. You could also argue that $\mathbb{R} = \bigcup_{i =1}^{\infty} C_{i} \cup \bigcup_{q \in \mathbb{Q}} \{q\}$, so by Baire one of the $C_{i}$ must have non-empty interior, contradicting the fact that $\mathbb{R} \smallsetminus \mathbb{Q}$ has empty interior.

Suppose $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then $\mathbb{Q}$ is a $G_{\delta}$ set which is a contradiction. In other words, suppose $$ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i $$ Then by Demorgan's Laws $$ \mathbb{Q} = \bigcap_{i=1}^{\infty} \ C_i^{c} $$ which is a contradiction since $\mathbb{R} \setminus \mathbb{Q}$ is a countable intersection of open sets and we know that the intersection between the rational is irrationals is $\emptyset$. So $\emptyset$ is a countable intersection of open dense subsets which contradicts Baire's category theorem.