According to the comment in my previous answer, let met try a more general approach. I assume that $V$ is a normed vector space such that the unit sphere has a unique supporting hyperplane for each of its boundary point.
Warning: this is just a preliminary work that I am posting not to lose it. I intend to finish it when I have more time...
The key point is to prove that larger and larger spheres ressemble more and more to half-spaces. A precise statement could be the following: Choose a unit vector $e \in V$ and define $B_\lambda = B(-\lambda e, \lambda)$ for all $\lambda > 0$.
- Claim 1: If $\lambda' \geq \lambda$, $B_\lambda \subset B_{\lambda'}$.
Indeed, if $x \in B_\lambda$, $\|x + \lambda e\| \leq \lambda$. Thus
$$
\|x+ \lambda' e \| = \|x+ \lambda e + (\lambda' - \lambda)e\| \leq
\|x+ \lambda e\| + (\lambda' - \lambda)\|e\| \leq \lambda',
$$
showing that $x \in B_{\lambda'}$.
- Claim 2: $C = \bigcup_{\lambda > 0} B_\lambda$ is a convex cone.
It is convex as if $x, y \in C$, there exists $\lambda > 0$ so that $x, y \in B_\lambda$, so the segment $[x,y]$ is in $B_\lambda \subset C$. Now, if $x \in C$ and $\alpha \geq 0$, let $\lambda > 0$ be such that $x \in B_\lambda$. Then we have
$$
\|\alpha x + \alpha \lambda e\| \leq \alpha \|x + \lambda e\| \leq \alpha \lambda
$$
so $\alpha x \in B_{\alpha \lambda} \subset C$.
It should be noted that $C$ is not closed. On $\mathbb{R}^n$, if we choose $e = e_1$, then $C = \{x \in \mathbb{R}^n, x_1 < 0\} \cup \{0\}$.
- Claim 3: $\overline{C}$ is a half-space.
As $e \not\in B_\lambda$ for all $\lambda$, $e \not\in C$ so, from then Hahn-Banach theorem, there exists $\phi \in V^*$ such that $\phi(e) > 0$, $\phi(x) \leq 0$ for all $x \in \overline{C}$. If $\overline{C} \neq \{x \in V, \phi(x) \leq 0\}$, there exists $x_0 \in V$ such that $\phi(x_0) < 0$ (as the complement of $\overline{C}$ is open). From the Hahn-Banach theorem, there exists $\psi \in V^*$ such that $\psi(x_0) > 0$ and $\psi(x) \leq 0$ for all $x \in \overline{C}$. $\phi$ and $\psi$ cannot be proportional so $\ker(\phi)$ and $\ker(\psi)$ are two distinct supporting hyperplanes for $\overline{C}$ and thus for (say) $B_1$. This contradicts the assumption showing that $\overline{C}^is a half-space.
Assume now that $A$ is a non-empty bounded convex set. As is well-known, $A$ is the intersection of all its supporting half-spaces, so we want to prove that $O(A)$ is contained in all these half-spaces. The idea is as in my previous answer: we construct a sequence of balls whose center goes to infinity in some direction so they have bigger and bigger radius and ressemble near infinity to the half-space we chose. Our first task is to find the right direction in which to move our centers:
- Claim 4: (assume that $V$ is reflexive) For any (vector) hyperplane $H$, there exists a vector $e \in \overline{B}(0, 1)$ such that $H$ is the supporting hyperplane of $\overline{B}(-e, 1)$ at $0$.
Let $\phi \in V^*$ be such that $H = \ker(\phi)$ and $\|\phi\|= 1$. If $\dim(V)$ is finite, there exists $e \in \overline{B}(0, 1)$ such that $\phi(e) = 1$. The affine hyperplace $\{x \in V, \phi(x) = 1\}$ is then trivially a supporting hyperplane for $\overline{B}(0, 1)$. The same conclusion holds if $V$ is reflexive as the unit ball is weakly compact and $\phi$ is weakly continuous.