I don't see why $W^{1, 2}(\partial D)$ being compactly embedded in $L^2(\partial D)$ lets us show an operator is Fredholm of index zero.

Question

Let $D$ be a bounded Lipschitz domain. Let $A$ be the single layer potential which maps $L^2(\partial D)$ into $W^{1, 2}(\partial D)$ boundedly. $A$ is given by:

$$ A_D[\phi] = \int_{\partial D}G(x-y)\phi(y) d\sigma(y), \quad x \in \mathbb{R}^3, $$

where $G$ is the fundamental solution of the Laplacian.

Now as $W^{1, 2}(\partial D)$ is compactly embedded in $L^2(\partial D)$ we have that $A$ is Fredholm of index zero.

I don't see how we can say this operator is Fredholm of index zero based on the given information? To show it is Fredholm of index zero we must show that $A$ is

1. a bounded linear operator
2. The range of $A$ is closed in $W^{1, 2}(\partial D)$.
3. The subspaces $\ker(A)$ and $\text{coker}(A)$ are finite-dimensional.
4. The dimensions of $\ker(A)$ and $\text{coker}(A)$ are the same.

We have point 1. by the definition of $A$, but I don't see how the statement

$W^{1, 2}(\partial D)$ is compactly embedded in $L^2(\partial D)$

gives us points 2. - 4? Just to mention I am unfamiliar with compact embedding but from reading up on definitions of it I don't see how it gives us points 2. - 4.?

Edit: $A$ is actually a single layer potential but I left that out to simplify the question. My question comes from the proof of Theorem 2.13 in this book on page 29.