Let $A\subset \mathbb R$ be any set and the closure of $A,$ that is, $\bar{A}$ is not compact in $\mathbb R.$ Then $\bar{A}$ must be unbounded in $\mathbb R.$
Let $X$ be an infinite dimensional Banach space with the norm $\|\cdot \|_{X}$, and $K$ be any subset of $X$ with the property that $\bar{K}$ is not compact in $X.$
My Question: What can we say about $\bar{K}$? Can we say there is a sequence $\{x_{n}\} \subset K $ and $\epsilon >0$ such that $\|x_n-x_{n'}\|_{X}\geq \epsilon$ for $n\neq n'$?
Note that $\overline{K}$ is complete, since it is a closed subset of a complete space. A metric space is compact iff it is complete and totally bounded, so $\overline{K}$ is not totally bounded. This means that for some $\epsilon>0$, for any finite subset $S\subseteq K$ there exists $x\in \overline{K}$ which is distance $>\epsilon$ from every point of $S$. We can then use this to construct a sequence $(y_n)$ of points in $\overline{K}$ by induction, where at each step we choose $y_n$ such that $\|y_n-y_m\|>\epsilon$ for all $m<n$. Now just choose $x_n\in K$ within $\epsilon/3$ of $y_n$ for each $n$, and the sequence $(x_n)$ satisfies $\|x_n-x_m\|>\epsilon/3$ for any distinct $m$ and $n$.