Sierpinski himself proved the following striking version of his theorem for the plane:
CH is equivalent to the statement that $\mathbb R^3$ can be written as $A_x\cup A_y\cup A_z$ where each $A_u$, $u\in\{x,y,z\}$, has a finite intersection with every line parallel to the $u$-axis.
(See http://www.math.wisc.edu/~miller/old/m873-05/setplane.ps, respectively [Simms, John C.; Sierpinski’s theorem. Simon Stevin 65 (1991), no. 1-2, 69–163]. I believe this theorem is proved in Sierpisnki's 1951 paper in Fund. Math. 38, pp. 1--13, but I would have to actually walk to the library to verify this (and decipher enough of the french in the article).)
There is actually a planar version of this result of Sierpinski, due to Komjath.
A set $A\subseteq\mathbb R^2$ is a cloud if there is $p\in\mathbb R^2$ such that every line through $p$ meets $A$ in only finitely many points.
CH is equivalent to $\mathbb R^2$ being the union of 3 clouds.
In both cases the striking fact is that we go down from countable to finite, which is somehow less arbitrary, in the light of Joel's comment that the previous constructions can be done in ZFC if countable is replaced by $\lle \lneq 2^{\aleph_0}$.
