Suppose $B(t)$ is a standard Brownian motion in $\mathbb{R}^2$ started from the origin. It is will known that almost surely, for any point $(x,y) \in \mathbb{R}^2$ (including (0,0)), $P(B(t) = (x,y) \mbox{for some t >0})=0$, i.e. 2-d Brownian motion does not hit any particular point with probability 1.
A proof of this fact is given in Corollary 2.23 in the excellent book Morters and Peres, here. This proof however seems somewhat over complicated! It relies on the fact that the image of Brownian motion in the plane has Lebesgue measure 0. Another intuitive reason as to why this must be true is that the two component brownian motions cannot both equal zero at the same time, since at any particular point they are independent, continuous random variables. I cannot though come up with a simple argument for why this must be true based on this fact.
My question then is: does anyone know of a simpler proof of this fact that is of the flavor "The two component brownian motions cannot be equal to zero at the same time, since at any particular point they are independent, continuous random variables?"