I'm doing Ex 3.25 in Brezis's book of Functional Analysis.
Let $K$ be a compact metric space that is not finite. Prove that $\mathcal C(K)$ is not reflexive.
In below attempt, I only use the compactness of $K$ to ensure that every real-valued function on $K$ is bounded. I did not use the metric structure nor the finiteness of $K$. I guess I made some mistake but could not recognize.
Could you shed some lights on where I got wrong?
Let $(y_m)$ be a sequence in $K$ such that $y_m \to y \in K$. Let $E := \{f \in \mathcal C(K) \mid f(y) =0\}$. Then $E$ is a closed subspace of $\mathcal C(K)$. Assume the contrary that $\mathcal C(K)$ is reflexive. Then $E$ is also reflexive. Consider $$ \varphi:E \to \mathbb R, f \mapsto \sum_{m\ge 1} \frac{f(y_m)}{2^m}. $$
Clearly, $\varphi$ is linear. Also, $$ |\varphi(g)-\varphi(f)| = \bigg | \sum_{m\ge 1} \frac{(g-f)(y_m)}{2^m} \bigg | \le \sum_{m\ge 1} \frac{|g-f|_\infty}{2^m} = |g-f|_\infty. $$
It follows that $\varphi$ is continuous and thus $\varphi \in E'$. It's easy to see that $\|\varphi\| = 1$. By Hahn-Banach theorem, there is $\Psi \in E''$ such that $\|\Psi\| =\Psi(\varphi) = 1$.
Because $E$ is reflexive, there is $h\in E$ such that $|h|_\infty = \|\Psi\|$ and $\Psi(\varphi) = \varphi(h)$. This implies $\sum_{m\ge 1} 2^{-m} h(y_m) = 1$. Notice that $\sum_{m\ge 1} 2^{-m} h(y_m) \le \sum_{m\ge 1} 2^{-m} |h(y_m)| \le |h|_\infty = 1$, so we get $h(y_m) = |h|_\infty = 1$ for all $m$. It follows from $y_m \to y$ that $h(y)=1 \neq 0$, which is a contradiction.
Because $K$ is infinite, we can pick the sequence $(y_m)$ such that $y_m \neq y$ for all $m$. Let $B_E$ be the closed unit ball of $E$. Notice that $\{y_1, \ldots, y_m, y\}$ is closed. Any metric space is normal. By Tietze extension theorem, there is $f_m \in \mathcal C(K)$ such that $f_m (y)=0$ and $f_m (y_1) = f_m (y_2) = \cdots = f_m(y_m)=1$. By construction, $f_m \in B_E$ for all $m$. Because $E$ is reflexive, there is a subsequence $(f_{m_n})_n$ and $g\in B_E$ such that $g_n :=f_{m_n} \rightharpoonup g$ in the weak topology of $E$.
Notice that the map $E \to \mathbb R, f \mapsto f(x)$ is linear continuous in supremum norm topology, so it is in the weak topology of $E$. It follows that $g_n(y_m) \to g(y_m)$ as $n \to \infty$. It follows that $g(y) = 0$ and $g(y_m) = 1$ for all $m$. Because $y_m \to y$ and $g$ is continuous, $1=g(y_m) \to g(y)$ and thus $g(y)=1$. This is a contradiction.