Let $A_n=\{x_n,x_{n+1},...\}\subset E$ for each $n\in \mathbb N$, such that $E$ is a Banach space.
If the closure of the convex hull of $A_n$ is compact i.e $\overline{co}(A_n)$ compact, is $\overline{co}(A_1)$ compact?
what we can say about $A_1$?
It is well-know that the closure of the convex-hull $\overline{\mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $\overline{A_n} \subset \overline{\mathrm{Co}(A_n)}$. Thus, if $\overline{\mathrm{Co}(A_n)}$ is compact, then $\overline{A_n}$ is compact too. Since $$A_1 \subset \{x_1,\ldots, x_{n-1}\} \cup \overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $\overline{\mathrm{Co}(A_1)}$ is compact. In fact, note that $$\overline{\mathrm{Co}(A_1)} \subset \overline{\mathrm{Co}(K)}$$ and the last set is compact (because $K$ is compact).