All set in the Borel $\sigma$-algebra of $(\mathbb{R}, \tau_{usual})$, there are only $F_{\sigma}$ and $G_{\delta}$

Question

Our teacher affirms this, but we have found a counterexample in the internet : $\mathbb{I}^{-} \cup \mathbb{Q}^{+} $ where:

$\mathbb{I}^{-}=\mathbb{I} \cap (-\infty,0) $ and $\mathbb{Q}^{+}=\mathbb{Q} \cap (0,\infty)$.

We know that $\mathbb{Q}$ is $F_{\sigma}$ (countable union of points, and points are closed) and the complementary, $\mathbb{I}$ is $G_{\delta}$ because it is the complementary of an $F_{\sigma}$ .

We need to prove that $\mathbb{I}^{-} \cup \mathbb{Q}^{+} $ isn`t an $F_{\sigma}$.

Do someone could help us?