In general it may fail to be compact. Consider $E:=L_2(\mathbb{R})$, and $x_n:=\chi_{[n,n+1]}$. Clearly, for any $n\in\mathbb{N}$, all functions in the set $\overline{\text{co}}\{x_k:k\ge n\}$ have support in $[n,+\infty)$, and in the intersection we get a compact nonempty set, the singleton $X_\infty=\{0\}$. However no subsequences of $x_n$ converges strongly, since it already converges weakly to $0$, and $\|x_n\|_2=1$.

In general it may fail to be compact. Consider $E:=L_2(\mathbb{R})$, and $x_n:=\chi_{[n,n+1]}$. Clearly, for any $n\in\mathbb{N}$, all functions in the set $\overline{\text{co}}\{x_k:k\ge n\}$ have support in $[n,+\infty)$, and in the intersection we get a compact nonempty set, the singleton $X_\infty=\{0\}$. However no subsequences of $x_n$ converges strongly, since it already converges weakly to $0$, and $\|x_n\|_2=1$.

In general, in any first countable topological space, the compactness of a sequence $(x_n)_n$ is equivalent to $\bigcap_{n\ge k}\overline{\{x_n:n\ge k\}}$ being non-empty, since the latter is the set $\text{Lim}(x_n)$ of the limits of all converging subsequences . In a Banach space, this set may be any separable closed subset.