Every path-connected space is connected. So I think it is easier to show that it is path-connected. The proof for path-connectedness of the above problem can be found here.
There's a slightly simpler argument than the one presented. Given any two points x and y in R^2 - C, there are uncountably many pairwise disjoint (except at the endpoints) paths from x to y. (E.g., take any point z on a given line perpendicular to that between x and y, and then draw the path from x to z to y.) Since there are only countably many points in C, at least one of these paths (actually, uncountably many) must be free of any of these points. Hence R^2 - C is path-connected.