Cut Points on Brownian Paths

Let $X$ be a standard two-dimensional Brownian motion. There exists a.s. $t \in (0, 1)$ such that $X(\lbrack 0, t)) \cap X((t, 1 \rbrack) = \varnothing$. It follows that $X(\lbrack 0, 1 \rbrack)$ is not homeomorphic to the Sierpinski carpet a.s.