I want prove that the Wiener algebra, $W(\mathbb{T})$, does not coincide with the disc algebra $A(\mathbb{D})$. I know that there are some ways to do this. For example, one smart way is use Rudin-Shapiro sequence, see this post.
Note that $W(\mathbb{T})$ is an isometry to $\ell^1$, with pre-dual $c$ (or $c_0$, even something else). However, I never hear story about the pre-dual of disc algebra.
Question: Does $A(\mathbb{D})$ has pre-dual? What it is?
As you noticed, the Wiener algebra is isometric to $\ell_1$. In particular, it has the Schur property (weakly convergent sequences converge in the norm). However, this is very much not the case for the disc algebra as it, for example contains a sequence equivalent to the standard basis of $c_0$. It is a nice exercise to find explicitly a sequence in the disk algebra that converges weakly but not in the norm.
However, since $A(\mathbb D)$ is separable and contains a copy of $c_0$, by Sobczyk's theorem, this copy is complemented. No dual space contains a complemented copy of $c_0$ (Theorem 2.4.15). Hence $A(\mathbb D)$ is not isomorphic to a dual space.
For more details, see this paper.