If $\operatorname{cf}(\alpha)=\omega$, $D$ can be empty even if $\lambda=2$. Take $\alpha=\omega$: $\{2n:n\in\omega\}$ and $\{2n+1:n\in\omega\}$ are disjoint cub’s in $\omega$. If you want an uncountable example, take $\alpha=\omega_\omega$; $\{\omega_n:n\in\omega\}$ and $\{\omega_n+1:n\in\omega\}$ are disjoint cub’s. (The last sentence before Definition 6.7 even points this out.)
The specific error in your argument is the claim that $\gamma\in C_\beta$ for every $\beta$: the fact that $\gamma$ is less than some member of $C_\beta$ does not imply that $\gamma\in C_\beta$: $C_\beta$ is in general not a transitive set.
Added: In the textbook argument you have a strictly increasing sequence $\langle g^n(\xi):n\in\omega\rangle$ of ordinals less than $\alpha$; since $\operatorname{cf}(\alpha)>\omega$, $g^\omega(\xi)\triangleq\sup\{g^n(\xi):n\in\omega\}<\alpha$ as well. Moreover, $g^0(\xi)=\xi$, so $g^\omega(\xi)>\xi$. Now consider the cub $C_\beta$. Recall that $f_\beta(\xi)$ is the smallest member of $C_\beta$ larger than $\xi$ and that $g(\xi)=\sup\{f_\gamma(\xi):\gamma<\lambda\}\ge f_\beta(\xi)$, so at this point we have $\xi Then we repeat the process: $g^1(\xi)=g(g^0(\xi))=g(g(\xi))$, so $(1)$ becomes $g(\xi) In general we have
$g^n(\xi) for each $n\in\omega$, so we have an infinite chain
$\xi=g^0(\xi) where $f_\beta(g^n(\xi))\in C_\beta$ for each $n\in\omega$. Since the strictly increasing sequences $\langle f_\beta(g^n(\xi)):n\in\omega\rangle$ and $\langle g^n(\xi):n\in\omega\rangle$ are ‘interlocked’, they have the same supremum, and therefore $g^\omega(\xi)=\sup\{f_\beta(g^n(\xi)):n\in\omega\}$. Again, recall that $f_\beta(g^n(\xi))\in C_\beta$ for each $n\in\omega$, so $g^\omega$ is the limit from below of members of $C_\beta$: $C_\beta\cap g^\omega(\xi)$ is an unbounded subset of $g^\omega(\xi)$, and therefore, by the definition of cub, $g^\omega\in C_\beta$. And since $\beta<\lambda$ was arbitrary, $g^\omega\in D$.