Proof of the bilinear analogue of Schauder's theorem

Question

I am trying to find a proof of the following statement, which is the bilinear analogue of Schauder's theorem for linear operators. I know that this is proved (I think for the first time) in the article ''Operator ideals and spaces of bilinear operators '' by Ramanujan and Schock, but the proof is very densely written making it difficult to follow. Is there a reference where I can find this proof in more detail or can someone give a detailed proof?

The statement

Let $A$,$B$ and $E$ be Banach spaces (over $\mathbb{C}$) and let $T: A \times B \rightarrow E$ be a bounded bilinear operator. Then, $T$ is a compact bilinear operator if and only if the adjoint operator of $T$, $T^\times:E^* \rightarrow \mathcal{L}(A \times B,\mathbb{C})$ is a compact linear operator.

Preliminary notions:

Let $A,B,E$ be normed spaces.

1. A bilinear operator $T:A \times B \rightarrow E$ is called bounded if there exists some finite constant $C>0$ such that for all $a \in A$ and $b \in B$, $\|T(a,b)\|_E \leq C \cdot\|a\|_A \cdot \|b\|_B$. We denote by $\boldsymbol{\mathcal{L}(A \times B,E)}$ the set of all bounded bilinear operators $T: A \times B \rightarrow E$.

2. Let $T \in \mathcal{L}(A \times B,E)$. The adjoint operator $T^\times$ of $T$ is the linear operator $T^{\times}:E^* \rightarrow \mathcal{L}(A \times B, \mathbb{C})$ defined by the formula $T^\times(f)= f \circ T$.

3. A bilinear operator $T:A \times B \rightarrow E$ is called compact if for every bounded $S \subset A \times B$, the closure of $T(S)$ is compact in $E$.