Let $X$ be a Banach space, and $A\subseteq X$ subgroup, where $A$ is not countable, and there is some $d \gt 0$ such that for all $x,y \in A$: $||x-y||>d$. Prove that $X$ is not separable.
My attempt -
Let's assume that $X$ is separable, and let $B$ be the dense countable subset of $X$. Than $A\subseteq \overline B$. Meaning that every element of $A$, is a limit point of some sequence in $B$.
So if I could find a Cauchy sequence in $B$, such that it's limit is in $A$ (that Cauchy sequence has a limit since $X$ is Banach), the property of elements in $A$(that every two elements are at least of distance $d$), would contradict the propeties of Cauchy sequences, and hence $X$ cannot be separable.
Two things that I havn't used are that $B$ is countable and that $A$ is not countable. How can I prove that there is at least one Cauchy sequence in $B$ that converges to some point in $A$?
In fact, how can I know at all that there is Cauchy sequence in $B$?
Or perhaps, is there any other way?
$X$ is separable iff it contains a dense countable subset. If $A$ if an additive subgroup with points separated more than $d$, then the open sets $(U_a)_{a \in A}$ given by $$ U_a = B_a(d/3), $$ Form an uncountable family of disjoint open sets. Any dense set must intersect each of them, but by disjointedness this would imply that the dense set has uncountable cardinality.