Maximum $n$-dimensional volume of set satisfying $(\forall x) \: (\forall y) \: (\| x - y \| \le 1)$ Let $S \subseteq \mathbb{R}^n$ be a set satisfying the property:
$$P(S) \equiv (\forall x \in S) \: (\forall y \in S) \: (\| x - y \| \le 1) $$
where $\| \cdot \|$ is the Euclidean norm.
Let $\mathcal{S}$ be a set of all maximal sets $S$ (maximal by set inclusion) s.t. $S$ satisfies the property $P(S)$. Then what is the value of $V = \max_{T \in \mathcal{S}} \operatorname{vol}(T) $ where $\operatorname{vol}(\cdot)$ represents the $n$-dimensional volume of a set.
I can't figure out the shape of the set for even $n=2$, is it a circle with radius $\frac{1}{2}$, or a square with side $\frac{1}{\sqrt{2}}$, I can't tell.

Let $S \subseteq \mathbb{R}^n$ satisfy:
$P(S) \equiv (\forall x \in S) \: (\forall y \in S) \: (\| x - y \|_2 \le 1) $. Define the notation $B_n(r,p)=$ the $n$-dimensional hyperball with radius $r$, centre $p$.
Let $M=\{S\subseteq\mathbb{R}^n: [P(S)=\top] \wedge [\nexists S'\in\mathbb{R}^n \text{ s.t. } (P(S')=\top) \wedge (S'\supsetneq S)]\}$. Claim: $V := \max_{T \in M} \operatorname{vol}(T) =\operatorname{vol}(B_n(1/2, 0))$.
Note: This problem seems intuitively easy but deceptively hard. I attach my wrong proof in the hope that someone can fix it. The $\color{red}{red}$ implication doesn't hold, so the proof is invalid.
$\color{red}{Wrong}$ Proof: Note that it is sufficient to prove that $M=A$, where we define $A:=\{B_n(1/2,p): p\in\mathbb{R}^n\}$. It is obvious that $A\subseteq M$. Now we try to prove $M\subseteq A$:
Given $S\in M$, so that $P(S)=\top$ and $S$ is maximal. Note that $P(S)=\top\color{red}{\nRightarrow} \exists$ $p$ such that there is a circumscribed $B_n(1/2,p)\supseteq S$. But maximality implies that there must be equality, so $S=B_n(1/2,p)$. Thus, $S\in A$. $\blacksquare$
