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