2011 AMC 10B Problem 20

The points closer to $B$ than to another vertex are bounded by the perpendicular bisectors of $BA$, $BC$, and $BD$. The bisector of $BD$ is diagonal $AC$, so the desired region lies in $\triangle ABC$. Triangle $ABC$ has area $\dfrac12\cdot2\cdot2\sin120^\circ=\sqrt3$. The perpendicular bisectors of $BA$ and $BC$ cut off two congruent $30$-$60$-$90$ triangles, each with area $\dfrac{\sqrt3}{6}$. Therefore the desired area is $\sqrt3-2\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}$. Thus, C is the correct answer.