Let $(X, d)$ be a metric space. Let $J$ be an indexing set. Consider a set of the form $S = \{x_j\in X\mid j\in J\}$ with the property that $$d(x_j, x_k) = 1$$for all $j\neq k$, $j, k \in J$.
Which of the following statements are true?
(a) If such a set exists in $X$, then there exist open sets $\{U_j\}$ in $X$ such that $U_j \cap U_k = \emptyset$, for all $j \neq k$.
(b) There exists such a set $S$ in $C[0, 1]$ with $J$ being uncountable.
My attempt:
Option a is correct. Consider the sets $U_j= B_\frac{1}{3}(x_j)$ which are open (singletons are open sets in discrete metric). Clearly $U_j \cap U_k=\emptyset$ as $d(x_j, x_k)=1$.
How to disprove option b? Is the explanation correct for option a?
$C [0,1]$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $\{f_n\}$ be dense and for each $j \in J$ pick $k_j$ such that $d(f_{k_j}, x_j) <\frac 1 2$. Now verify that $j\in J \to k_j$ is a one to one map from $j$ into the set of positive integers. Hence $J$ is necessarily countable.