Set $J = (0,1)$, $K_1 = (0, \tfrac{1}{\mu})$, and $K_2 = (1 - \tfrac{1}{\mu}, 1)$. Since $\mu > 2$, we have $\tfrac{1}{\mu} < \tfrac{1}{2} < 1 - \tfrac{1}{\mu}$, so $K_1$ and $K_2$ are disjoint. On $K_1$: $F_\mu(x) = \mu x$, which maps $(0, \tfrac{1}{\mu})$ onto $(0,1) = J$. On $K_2$: $F_\mu(x) = \mu(1-x)$, which maps $(1 - \tfrac{1}{\mu}, 1)$ onto $(0,1) = J$. Hence $F_\mu(K_1) = F_\mu(K_2) = J$ and $K_1, K_2 \subset J$, giving a horseshoe for $F_\mu$.