Coloring $\mathbb{R}^2$ and single-colored paths Is there a colouring of $\mathbb{R}^2$ with $2$ colours, such that the just single-coloured paths are constant ones? (Note: I've got this problem from another one.)
 A: Yes, this is possible using the axiom of choice.  The key observation is that the image of any nonconstant path is a closed subset of $\mathbb{R}^2$ of cardinality $\mathfrak{c}$, and there are only $\mathfrak{c}$ such closed subsets.  You can then construct a coloring by transfinite induction so that each such set has points of both colors.
In detail, let $(X_\alpha)_{\alpha<\mathfrak{c}}$ be an enumeration of all the closed subsets of $\mathbb{R}^2$ of cardinality $\mathfrak{c}$.  We define sequences $(r_\alpha)_{\alpha<\mathfrak{c}}$ and $(b_\alpha)_{\alpha<\mathfrak{c}}$ by induction.  Having defined $r_\beta$ and $b_\beta$ for all $\beta<\alpha$, define $r_\alpha$ and $b_\alpha$ to be two distinct points of $X_\alpha$ which are not equal to $r_\beta$ or $b_\beta$ for any $\beta<\alpha$.  This is possible since we have chosen fewer than $\mathfrak{c}$ points so far, and $X_\alpha$ has cardinality $\mathfrak{c}$.
We thus obtain two disjoint sets $R=\{r_\alpha\}_{\alpha<\mathfrak{c}}$ and $B=\{b_\alpha\}_{\alpha<\mathfrak{c}}$ which each intersect every $X_\alpha$.  Color all the points in $R$ red, and all the points in $B$ blue, and all the points that are in neither $R$ nor $B$ however you want.  Then neither the red points nor the blue points contain any $X_\alpha$, and thus neither contains the image of a nonconstant path.
[This answer is adapted from my answer to Infinite 'hex': is a win always achievable?.  This construction (known as "Bernstein sets") has many variants and is useful as counterexamples to many questions of this sort.]
