I've just written what seems to me like a proof that weak$^*$ convergence implies uniform convergence in a reflexive Banach space, but I'm pretty sure that this can't possibly be true. Could somebody please point out the flaw in the following proof?
We let $X$ be a reflexive Banach space, $X^*$ be its continuous dual, and $\{f_n\}_{n=1}^{\infty}\subset X^*$ be a sequence that converges weak$^*$ to $f$ (which is to say, $f_n(x)\to f(x)$ for all $x\in X$). As $X$ is reflexive, the closed unit ball $\overline{B} = \overline{B_1(0)}\subset X$ is weakly compact. For $\epsilon > 0$, we let $\mathcal{F}_{\epsilon} : X\to \mathbb{N}$ be defined by $$\mathcal{F}_{\epsilon}(x) := \sup\{n\in \mathbb{N} : \lvert (f-f_n)(x)\rvert > \epsilon\}$$ and we let $\mathcal{G} : \mathbb{R}\to \mathbb{N}\cup \{\infty\}$ be defined by $$\mathcal{G}(\epsilon) := \sup_{x\in \overline{B}} \mathcal{F}_{\epsilon}(x)$$ If $\mathcal{G}(\epsilon) < \infty$ for all $\epsilon > 0$, then we have that for $\epsilon > 0$, $\|f-f_n\| < \epsilon$ for $n > \mathcal{G}(\epsilon)$, and therefore, $\|f-f_n\|\to 0$. If $\mathcal{G}(\epsilon) = \infty$ for some $\epsilon > 0$, then we have some sequence $\{x_i\}_{i=1}^{\infty}\subset \overline{B}$ such that $\mathcal{F}_{\epsilon}(x_i)\to \infty$. As $\overline{B}$ is weakly compact, we have some subnet $\langle x_{\alpha}\rangle_{\alpha\in \mathcal{A}}$ of $\langle x_i\rangle_{i\in \mathbb{N}}$ such that
(1) $\mathcal{F}_{\epsilon}(x_{\alpha})\to \infty$, and
(2) $\langle x_{\alpha}\rangle_{\alpha\in \mathcal{A}}$ converges weakly to some $x\in \overline{B}$, which is to say that $\langle g(x_{\alpha})\rangle_{\alpha\in \mathcal{A}}$ converges to $g(x)$ for all $g\in X^*$.
Therefore, we have that $\langle (f-f_n)(x_{\alpha})\rangle_{\alpha\in \mathcal{A}}$ converges to $(f-f_n)(x)$ for all $n$. For $N\in \mathbb{N}$, there is an $\alpha'\in \mathcal{A}$ such that for all $\alpha\geq \alpha'$ in $\mathcal{A}$, $\mathcal{F}_{\epsilon}(x_{\alpha}) > N$, which implies that $\lvert (f-f_n)(x_{\alpha})\rvert > \epsilon$ for some $n > N$ for all $\alpha\geq \alpha'$. But, as $(f-f_n)(x_{\alpha})$ converges to $(f-f_n)(x)$, we have some $\alpha''$ such that $\lvert (f-f_n)(x-x_{\alpha})\rvert < \frac{\epsilon}{2}$ for all $\alpha\geq \alpha''$. Therefore, for $\alpha\geq \max(\alpha', \alpha'')$, $\lvert (f-f_n)(x-x_{\alpha})\rvert < \frac{\epsilon}{2}$ and $\lvert (f-f_n)(x_{\alpha})\rvert > \epsilon$, so $\lvert (f-f_n)(x)\rvert > \frac{\epsilon}{2}$ for some $n > N$. As $N$ can be arbitrarily large, we have that $\lvert (f-f_n)(x)\rvert\not\to 0$, which contradicts that $f_n$ converges weak$^*$ to $f$. Therefore, $\mathcal{G}(\epsilon) < \infty$ for all $\epsilon > 0$, so $\|f-f_n\|\to 0$.
Thanks for your help!