The statement is false in any unit sphere of an infinite dimensional Banach space. A compactness argument together with the form of Dvoretsky's theorem stated here (theorem 1.2), gives:

**Proposition 1:** For any $\varepsilon > 0$ and $n<\omega$ there is a $K(\varepsilon, n)$ such that, for any Banach space $X$ of dimension $\geq K(\varepsilon ,n)$ with unit sphere $S$ and any $1$-Lipschitz $f: S \rightarrow [0,2]$, there is an $n$-dimensional subspace $Y \subseteq X$ such that for all $x,y \in Y\cap S$, $|f(x)-f(y)| < \varepsilon $.

*Proof:* Assume that the statement is false for some $\varepsilon >0$ and $n<\omega$. Then for every sufficiently large $k$ we can find a Banach space $X_k$, with unit sphere $S_k$, and a $1$-Lipschitz function $f_k : S_k \rightarrow [0,2]$ such that for every $n$ dimensional subspace $Y$ of $X_k$, there are $x,y \in Y \cap S_k$ such that $|f(x)-f(y)|\geq \varepsilon$. Take an ultraproduct of the sequence $(X_k, f_k)$ to get $(X,f)$, with $X$ the Banach space ultraproduct and $f$ defined in the obivous way. You can check that $f$ is well-defined, $1$-Lipschitz, and takes values in $[0,2]$. It will also be infinite dimensional and therefore give a counterexample to theorem 1.2 in the reference (a certain form of Dvoretzky's theorem). $\square$

($[0,2]$ is just the range of the metric when restricted to the unit sphere of an infinite dimensional Banach space.)

**Proposition 2:** Let $S$ be the unit sphere of an infinite dimensional Banach space $X$ and let $sS$ be its Samuel compactification. There is an $x \in sS$ that is not in the closure of any uniformly discrete subset of $S$.

*Proof:* Let $D\subset S$ be a uniformly discrete subset that is $(>\varepsilon)$-separated. Assume without loss that $\varepsilon < 1$. By proposition 1, for any $n\geq 2$, for any subspace $Y\subseteq X$ of dimension at least $K(\frac{\varepsilon}{5}, n)$, there is a subspace $Z \subseteq Y$ of dimension $n$ such that for any $x,y \in Z \cap S$, $|d(x,D)-d(y,D)|< \frac{\varepsilon}{5}$.

Now suppose for the sake of contradiction that there is $x \in Z \cap S$ such that $d(x,D) < \frac{\varepsilon}{5}$. Let $u \in D$ be such that $d(x,u) < \frac{\varepsilon}{5}$. By the reverse triangle inequality, $d(-x,u) > 2-\frac{\varepsilon}{5} > \frac{\varepsilon}{2}$, so since $n\geq 2$, $Z \cap S$ is connected and thus there is a $y \in Z \cap S$ such that $d(y,u) = \frac{\varepsilon}{2}$. By construction $d(y,D) < \frac{\varepsilon}{5} + \frac{\varepsilon}{5}$, so there must be some $v\in D \setminus \{u\}$ such that $d(y,v) < \frac{2\varepsilon}{5}$, but this implies that $d(u,v) < \frac{\varepsilon}{2} + \frac{2\varepsilon}{5} < \varepsilon$, which contradicts that $D$ is $(>\varepsilon)$-separated. Therefore $d(x,D) \geq \frac{\varepsilon}{5}$ for every $x p\in Z \cap S$.

Therefore in particular, for any $(>\varepsilon)$-separated set $D \subset S$, there are arbitrarily large subspaces $Y \subseteq X$ such that $Y\cap S \subseteq \{x \in S : d(x,D) \geq \frac{\varepsilon}{5}\}$.

This implies, again by proposition $1$, that for any finite sequence $D_0,D_1,\dots,D_{m-1}$ uniformly discrete sets which are $(>\varepsilon_0)$-, $(>\varepsilon_1)$-, ..., and $(>\varepsilon_{m-1})$-separated, respectively, the set $\bigcap_{i<m} \{x \in S : d(x,D_i) \geq \frac{\varepsilon_i}{5}\}$ is non-empty.

In the Samuel compactification $sS$, for any uniformly discrete set $D$, the function $d(x,D)$ has a unique continuous extension $f_D$ to $sS$ and in particular the closure $\overline{D}$ is precisely the zeroset of the function $f_D$.

So now for the same sequence of uniformly discrete sets as above, consider the closed set $F = \bigcap_{i<m} \{x \in sS : f_{D_i}(x) \geq \frac{\varepsilon_i}{5}\}$. We already showed that this set is non-empty and by the above statement about $\overline{D}$, $F$ is disjoint from each of the sets $\overline{D}$.

By compactness of $sS$, this implies that the intersection $\bigcap\{\{x \in sS : f_{D}(x) \geq \frac{\varepsilon}{5}\}:D\subset S \text{ }(>\varepsilon)\text{-separated} \}$ is non-empty and so contains some $x$ such that for any $(>\varepsilon)$-separated set $D \subset S$, $f_{D}(x) \geq \frac{\varepsilon}{5}$, so in particular $x \notin \overline{D}$. $\square$