Consider the standard embedding of the unit interval in $\mathbb R^2$ viz. $I=[0,1]\times \{0\} \subset \mathbb R^2$. Let $C$ denote the Cantor subset $C \subset I$ and define $U= \mathbb R^2 - C$, an open subset of $\mathbb R^2$.

I seem to remember that $\pi_1(U)$ has cardinality at least the continuum and so the fibers of the universal covering $\tilde{U} \to U$ are such big discrete sets that I would guess that $\tilde{U} $ can't be embedded in $\mathbb R^2$.

Consider the standard embedding of the unit interval in $\mathbb R^2$ viz. $I=[0,1]\times \{0\} \subset \mathbb R^2$. Let $C$ denote the Cantor subset $C \subset I$ and define $U= \mathbb R^2 - C$, an open subset of $\mathbb R^2$.

I seem to remember that $\pi_1(U)$ has cardinality at least the continuum and so the fibers of the universal covering $\tilde{U} \to U$ are such big discrete sets that I would guess that $\tilde{U} $ can't be embedded in $\mathbb R^2$.

EDIT Thanks to Petya and Ryan for explaining that $\pi_1(U)$ is actually countable and that what "I seem to remember" is false. Sincere apologies to all for my misleading answer.

For the sake of atonement, here is another argument for the countability of $\pi_1(U)$. Since $U$ is locally connected, locally compact and second countable, any connected covering (or even étalé space) of $U$ is second countable by the theorem of Poincaré-Volterra. Hence the fibers of the covering, being discrete, are countable. But these fibers are equipotent to $\pi_1(U)$ , which must thus be countable. This argument seems to be valid for any open subset of $\mathbb R^n$.