Question
Let $X$ be a Reflexive space. Suppose $(x_n)_n \in X$ is a bounded sequence and define $K_n=\overline{conv\{x_m,m\geq n\}}$. Then $x_n\overset{w}{\to}x_0$ if and only if $\cap_{n\in\mathbb{N}}K_n=\{x_0\}$.
So I know that $K_n=\overline{conv\{x_m,m\geq n\}}=\overline{conv\{x_m,m\geq n\}}^w$ by Mazur's Theorem and thus $x_0\in K_n$ for every $n\in\mathbb{N}$. Meanwhile the $K_n$'s form a decreasing sequence so if I could show that $diam K_n \to 0$ I could just use Cantor's Theorem. I don't suspect something like this actually happens however and I am not sure how to use the reflexive property.
Suppose $x_n \overset{w}\to x$. Take $y \in \cap_{n \ge 1} K_n$. Fix $l \in X^*$. Take $\epsilon > 0$. Take $N \ge 1$ so that $|l(x_n)-l(x_0)| \le \epsilon$ for all $n \ge N$. Note $y\in K_N$ implies that there exist $k \ge 1$, $N_1,\dots,N_k \ge N$, and $\lambda_1,\dots,\lambda_k$ with $\sum_{i=1}^k\lambda_i=1$ and $\|\sum_{i=1}^k\lambda_ix_{N_i}-y\|<\frac{\epsilon}{||l||}$. Then $$\left|\ell(y)-\ell(x_0)\right|\leq \left|\ell(y)-\ell(\sum_{i=1}^k\lambda_ix_{N_i})\right|+\left|\ell(\sum_{i=1}^k\lambda_ix_{N_i})-\ell(x_0)\right|$$ $$\leq\epsilon+\left|\sum_{i=1}^k\lambda_i\ell(x_{N_i})-\sum_{i=1}^k\lambda_i\ell(x_0)\right|$$ $$\leq\epsilon+\sum_{i=1}^k\lambda_i\left|\ell(x_{N_i})-\ell(x_0)\right|\leq 2\epsilon,$$ the last holding since each $N_i\geq N$. As this inequality holds for any $\epsilon>0$, we have that $\ell(y)=\ell(x_0)$ for all $\ell\in X^*$. And since $X^*$ separates points, we see $y=x_0$.
For the "if" direction, suppose $x_n \not\overset{w}\to x$. Since $(x_n)_n$ is bounded and $X$ is reflexive, there is some $y \not = x$ so that $x_{n_k} \overset{w} \to y$ for some subsequence of $(x_n)_n$. But, once again, as you've already shown, $y \in \cap_{n \ge 1} K_n$.