Can we say that : $ (A-B)\cap\overline{B}(0,r)\text{ is weakly compact, }\forall r>0 $ Let $X$ be a separable Banach space and $A,B$ are closed convex subsets of $X$ such that $B\subset A$ and 
$$
A\cap\overline{B}(0,r) \text{ and } B\cap\overline{B}(0,r) \text{ are weakly compact, } \forall r>0.
$$
Can we say that : 
$$
(A-B)\cap\overline{B}(0,r)\text{ is weakly compact, }\forall r>0
$$
with $A-B=\{a-b:a\in A,b\in B\}$.
 A: Doesn't this already fail in finite dimensions? Take $X = \mathbb{R}^2$ and let $A = B =$
the closed region bounded by $y = \sqrt{x^2 + 1}$ and $y = x + 1$, both for $x \geq 0$. The intersection with any ball is closed and therefore compact, but $A - B$ is not closed: it contains points arbitrarily close to $(0, -1)$ but does not contain that point itself.
A: This is not an answer, but I put it here to stop close votes that are apparently based on a misreading of the question.
I do not know the answer to the OP's question, but I will point out that both hypotheses are necessary. For example, drop the hypothesis that $A$ is convex. Then there is a counterexample with $A=B$; namely, $A=\bigcup_n \{(n+1)e_n, ne_n\}$, where $(e_n)$ is the unit vector basis of $\ell_1$. Bounded subsets of $A$ are finite and $A-A$ contains $(e_n)$, which is not preweakly compact.
Secondly, drop the hypothesis that $B\subset A$ but keep the convexity hypotheses. Let $X= C[0,1]$ ($X$ could be any separable non reflexive space that contains an infinite dimensional reflexive subspace). It is known that there are quasi complementary subspaces $X_1$ and $X_2$ that are both isomorphic to $\ell_2$. This means that $X_1 \cap X_2 =\{0\}$ and $X_1 - X_2$ is norm dense in $X$, and hence $(X_1 - X_2) \cap B(0,1)$ is even norm dense in the non weakly compact unit ball of $X$.
