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?

Define the "conflict range" of $x$ to be $d(x)=\min\{|x-y| : y\in \cal{A}(x)\}$.  That is, $d(x)$ measures the distance from $x$ to its (finite) set of conflicts; note that $d(x)>0$ for each $x$.  Choose a distance $\Delta>0$ such that there are uncountably many $x$ with $d(x) > \Delta$.  For instance, since$$d^{-1}((1,\infty)) \cup \bigcup_{k=0}^{\infty}d^{-1}\left((2^{-(k+1)},2^{-k}]\right)= \mathbb{R},$$
at least one of the sets in the union is uncountable, and we can choose $\Delta$ to be the lower endpoint of its argument.  Now let ${\cal{D}}=d^{-1}\left((\Delta,\infty)\right)$, and note that
$$
\bigcup_{k=-\infty}^{\infty}{\cal{D}}\cap(k\Delta, (k+1)\Delta]={\cal D}.
$$
Each of the sets ${\cal{D}}\cap(k\Delta,(k+1)\Delta]$ is conflict-free by construction (its diameter is less than any of its elements' conflict ranges), and at least one of them must be uncountable (since their union is uncountable). That set satisfies the conditions of the problem. $\square$
