The Verdier dual of a complex of sheaves $F$ on a complex manifold $X$ is defined in the derived category $D_b(X)$ as $\mathcal{RHom}(F,\mathbb{D}_X)$, where $\mathbb{D}_X$ is the dualizing "sheaf" (actually a member of the derived category). I have seen $\mathbb{D}_X$ be given as $\pi^! \mathbb{C}_{pt}$, where $\pi$ is the map from $X$ to a singleton, and for the case of $X$ a compact, complex manifold, $\mathbb{D}_X$ is the orientation sheaf, shifted in $D_b(X)$ by the dimension of $X$.
Now, consider the case that $X$ is compact. Then $\pi$ is a proper map, and $\pi_\ast = \pi_!$. Then, because adjoints of the same functor are naturally isomorphic to each other, we have that the adjoints of $R\pi_\ast$ and $R\pi_!$ are the same, i.e. $\pi^\ast = \pi^!$. But then $\pi^\ast \mathbb{C}_{pt} = \pi^! \mathbb{C}_{pt}$. The first of these is just the constant sheaf on $X$, while the second is supposedly $\mathbb{D}_X$. However, these are supposedly not quasi-isomorphic; $\mathbb{D}_X$ is supposed to be non-zero in degree n only, while $\mathbb{C}_{pt}$ is non-zero in degree 0 only.
What am I missing? I am guessing my mistake is because I do not understand the point of the dualizing sheaf. Can anyone explain why we use the dualizing sheaf instead of the constant sheaf?