Let $X$ be a separable Banach space and by $ w$ we shall indicate the weak topology on $X$.
Let $\mathcal{P}_{wkc}(X)$ be the collection of all nonempty $w$-compact convex subsets of $X$.
Let $\{C_n\}$ be a sequence in $\mathcal{P}_{wkc}(X)$. Then : $$ \cap_{p}\overline{\text{co}}\cup_{m\geq p}\frac{1}{m}\sum_{n=1}^{m}{C_n}\subset \cap_{p}\overline{\text{co}}\cup_{m\geq p}{C_m} $$
with: $\overline{\text{co}}(A)=\overline{\left\{\sum _{i=1}^{n}\lambda _{i}x_{i}:n\in \mathbb {N} ,\,x_{i}\in A,\,\sum _{i=1}^{n}\lambda _{i}= 1\right\}}.$
Proof:
For every $p \geq 1$ $$ \overline{\text{co}}\cup_{m\geq p}{\big(\frac{1}{m}C_1+ \frac{1}{m}\sum_{n=2}^{m}{C_n}\big) }\subset \overline{\text{co}}\cup_{m\geq p}{\frac{1}{m}C_1}+\overline{\text{co}}\cup_{m\geq p}{ \frac{1}{m}\sum_{n=2}^{m}{C_n}}\qquad (*) $$ The right-hand side is closed, because $\overline{\text{co}}\cup_{m\geq p}{\frac{1}{m}C_1}$ is $w$-compact. Upon taking the intersection over all $p$ one thus obtains $$ \cap_p\overline{\text{co}}\cup_{m\geq p}{\big(\frac{1}{m}C_1+ \frac{1}{m}\sum_{n=2}^{m}{C_n}\big) }\subset \cap_p\bigg( \overline{\text{co}}\cup_{m\geq p}{\frac{1}{m}C_1}+\overline{\text{co}}\cup_{m\geq p}{ \frac{1}{m}\sum_{n=2}^{m}{C_n}}\bigg) $$ By the following lemma:
Let $\{C_n\}$ consist of $w$-compact and $\{D_n\}$ of $w$-closed subsets of $X$. Assume that $\{C_n\}$ and $\{D_n\}$ are both nonincreasing. Then: $$\cap_{n}{ (C_n + D_n)} = \cap_{n}{C_n} + \cap_{n}{D_n}.$$
this gives $$ \cap_p\overline{\text{co}}\cup_{m\geq p}{\big(\frac{1}{m}C_1+ \frac{1}{m}\sum_{n=2}^{m}{C_n}\big) }\subset \cap_p\bigg( \overline{\text{co}}\cup_{m\geq p}{\frac{1}{m}C_1}\bigg)+\cap_p\bigg(\overline{\text{co}}\cup_{m\geq p}{ \frac{1}{m}\sum_{n=2}^{m}{C_n}}\bigg)\qquad (**) $$
It is easy to check that $\cap_p\bigg( \overline{\text{co}}\cup_{m\geq p}{\frac{1}{m}C_1}\bigg)$ is actually identical to $\text{co}\big(\{0\}\cup \frac{1}{p}C_1\big)$ (note that $\{0\}\cup \frac{1}{p}C_1$, being $w$-compact, has a convex hull that is also $w$-compact). Therefore, it follows immediately that $\cap_p\bigg( \overline{\text{co}}\cup_{m\geq p}{\frac{1}{m}C_1}\bigg)$ is equal to $\{0\}$, in view of the boundedness of $C_1$. Substitution in $(**)$ gives $$ \cap_p\overline{\text{co}}\cup_{m\geq p}{\big(\frac{1}{m}C_1+ \frac{1}{m}\sum_{n=2}^{m}{C_n}\big) }\subset \cap_p\bigg(\overline{\text{co}}\cup_{m\geq p}{ \frac{1}{m}\sum_{n=2}^{m}{C_n}}\bigg)\subset \cap_p\bigg(\overline{\text{co}}\cup_{m\geq p}{\big(\overline{\text{co}}\cup_{n=2}^{m}{C_n}\big)}\bigg) $$ which leads to $$ \cap_p\overline{\text{co}}\cup_{m\geq p}{\big(\frac{1}{m}C_1+ \frac{1}{m}\sum_{n=2}^{m}{C_n}\big) }\subset \cap_p\bigg(\overline{\text{co}}\cup_{n\geq 2}{C_n}\bigg)=\overline{\text{co}}\cup_{n\geq 2}{C_n} $$ The proof is now easily completed by induction.
My problem: Why we have $(*)$?
Since the closure of a convex set in a Banach space is convex, the right-hand side $R$ of ($*$) is convex. Since it is also closed, to show ($*$) it suffices to remark that $R$ contains a set $\frac{1}{m}C_1+ \frac{1}{m}\sum_{n=2}^{m}{C_n}$ for each $m\ge p$.