Lemma. Let $(X,\tau)$ be a topological space and $x\in X$. Let $\mathcal{B}\subseteq \tau$ be closed under finite intersections and let $x\in U, \forall U\in\mathcal{B}$. Then $\mathcal{B}$ is a local basis for $x$ if and only if for any net $(x_{\alpha})_{\alpha}$ in $X$ we have $$ x_{\alpha}\to x \iff (\forall U\in \mathcal{B})(\exists \alpha_0)(\forall \alpha\ge \alpha_0)(x_{\alpha}\in U).$$
Theorem. If $X$ is a separable normed space, then $(\overline{B}_{X^*}(0,1),\tau_{pt})$ is a first-countable space. ($\tau_{pt}$ is the topology of pointwise convergence).
Proof: Let $\phi\in B^* := \overline{B}_{X^*}(0,1)$. Consider $\{x_n:n\in\mathbb{N}\}\subseteq X$ countable and dense. Define $$ U_n := \{\psi\in B^*: |(\psi-\phi)(x_1)|<1/n\,\, \& \dots \&\,\,|(\psi-\phi)(x_n)|<1/n\}.$$ We show that $\{U_n: n\in\mathbb{N}\}$ is a local $\tau_{pt}$-basis for $\phi$. ...
Now the proof proceeds to use the lemma mentioned above, with $\mathcal{B} = \{U_n: n\in\mathbb{N}\}$. Obviously, $\phi\in U_n, \forall n$ and $\mathcal{B}$ is closed under finite intersections. However, I don't see why $U_n\in \tau_{pt}, \forall n$.
Any ideas? Thanks.
$U_n$ is open as the finite intersection $\bigcap_{i=1}^n e_{x_i}^{-1}[(\phi(x_i)-\frac1n, \phi(x_i)+\frac1n)]$, where $e_x: V^\ast \to \Bbb R$ is the map sending $\psi \in B^\ast$ to $\psi(x) \in \Bbb R$. As $\tau_{pt}$ makes these $e_x, x \in X$ continuous, $U_n$ must be open in that topology.