Existence of a certain subset of $\mathbb{R}$ 
To every real $x$ assign a finite set $\mathcal{A}(x)\subset \mathbb{R}$
  where $x\not\in \mathcal{A}(x)$. Does there exist $\mathcal{W}\subset
 \mathbb{R}$ such that:
$$1.\;\;\mathcal{W}\cap \mathcal{A}(\mathcal{W})=\varnothing\qquad
 2.\;\;|\mathcal{W}|=|\mathbb{R}|$$

This interesting problem was given to me by a friend, but I can't do it. Any ideas?

Let $\mathcal{Q}=\{[p,q]:\;p,q\in\mathbb{Q},\;p<q\}$. $\mathbb{Q}$ dense in $\mathbb{R}$ and $\mathcal{A}$ finite $\Rightarrow$ we may choose $\phi:\;\mathbb{R}\to\mathcal{Q}:$ $$(\text{i}):\;\;x\in\phi(x)\qquad (\text{ii}):\;\;\phi(x)\cap\mathcal{A}(x)=\varnothing$$
Since $|\mathcal{Q}|=|\mathbb{N}|$ there exists $I\in\mathcal{Q}$ such that $\text{card}\,\{x\in\mathbb{R}:\;\phi(x)=I\}=\mathfrak{c}\;(\Leftarrow$ König's th.$)$. Let $\mathcal{W}=\{x\in\mathbb{R}:\;\phi(x)=I\}$ and check $\mathcal{W}\cap\mathcal{A}(\mathcal{W})=\varnothing\;(\Leftarrow\mathcal{W}\subset  I$ and $I\cap \mathcal{A}(\mathcal{W})=\varnothing)$
