 Solution

Let $z_0$ be attracting and let $U$ be its immediate basin. If $z_0$ itself is critical there is nothing to prove. Otherwise suppose $U$ contains no critical point. The restriction $f:U\to U$ is then an unbranched covering. Since the complement of $U$ contains the Julia set and hence at least three points, $U$ is hyperbolic. Lift the covering to the universal cover $\mathbb{D}\to U$, choosing a lift that fixes a point above $z_0$. Because both maps are universal coverings, the lift is an automorphism of $\mathbb{D}$. A disc automorphism fixing an interior point has derivative of hyperbolic norm one there, whereas the multiplier at $z_0$ has modulus strictly below one. This contradiction proves that $U$ contains a critical point, whose orbit converges to $z_0$. Thus every attracting fixed point attracts a critical point.