Q:

Complement of a meagre subset of $\mathbb{R}$ contains an uncountable $G_\delta$ set

I'm trying to show that the complement of a meagre set $A \subseteq \mathbb{R}$ contains an uncountable $G_\delta$ set. 
Here is what I got so far : since $A$ is meagre, there exists nowhere dense sets $A_n$ such that
\begin{equation} A = \bigcup\limits_{n = 1}^{\infty}A_n.\end{equation}
The set $\bigcup\limits_{n = 1}^{\infty}\overline{A_n}$ is a $F_\sigma$ set of first category containing $A$. Therefore,
\begin{equation} \left(\bigcup\limits_{n = 1}^{\infty}\overline{A_n}\right)^c =\bigcap\limits_{n = 1}^{\infty}\overline{A_n}^c.\end{equation}
is a $G_\delta$ set contained in the complement of $A$.
I don't know how to show that this $G_\delta$ is uncountable. Does anyone have a clue ?

A:

HINT: Try to embed the Cantor set in $\Bbb R\setminus A$ by imitating the construction of the middle-thirds Cantor set. If you get stuck, this answer to an earlier question gives the argument in fairly condensed form, and this answer to an even earlier question gives more detail. (These are both done in an arbitrary complete metric space with no isolated points rather than specifically in $\Bbb R$, but that shouldn’t cause any great difficulty.)

