Theorem: Banach Space with Weak Topology has Countable Tightness

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,}_X}$.
Then $\struct {X, w}$ has countable tightness.

Proof:
Let $A \subseteq X$. 
Let $\cl_w$ be the closure taken in $\struct {X, w}$. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$. 
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation. 
Let $B_{X^\ast}^-$ be the closed unit ball in $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Let $x \in \map {\cl_w} A$. 
Fix $k, m \in \N$. 
Let $\FF_k$ be the set of $k$-tuples $\tuple {f_1, \ldots, f_k} \in \paren {B_{X^\ast} }^k$.
Let $F = \tuple {f_1, \ldots, f_k} \in \FF_k$.
From Open Sets in Weak Topology of Topological Vector Space:
:$\ds \set {x \in X : \cmod {\map {f_i} {x - x_0} } < \frac 1 m \text { for all } 1 \le i \le k}$ is open in $\struct {X, w}$.
Since $x \in \map {\cl_w} A$, there exists $\ds s := s_{F, k, m} \in A \cap \set {x \in X : \cmod {\map {f_i} {x - x_0} } < \frac 1 m \text { for all } 1 \le i \le k}$.
From Characterization of Continuity of Linear Functional in Weak-* Topology, $\paren {\iota s - \iota x_0} : X^\ast \to \GF$ is weak-$\ast$ continuous.
Since:
:$\ds \cmod {\map {\paren {\iota s - \iota x_0} } {f_i} } < \frac 1 m$ for all $1 \le i \le k$
there exists an open neighborhood $V_{F, i, k, m}$ of $f_i$ in $\struct {X^\ast, w^\ast}$ such that:
:$\ds \cmod {\map {\paren {\iota s - \iota x_0} } g} = \cmod {\map g {s_{F, k, m} - x_0} } < \frac 1 m$ for all $1 \le i \le k$ and $g \in V_{F, i, k, m}$.
Let:
:$\ds V_{F, k, m} = \prod_{i \mathop = 1}^k V_{F, i, k, m}$
From Natural Basis of Product Topology, $V_{F, k, m}$ is an open neighborhood of $\tuple {f_1, \ldots, f_k}$ in $\struct {\paren {B_{X^\ast} }^k, \paren {w^\ast}^k}$, where $\paren {w^\ast}^k$ denotes the product topology on $\struct {X^\ast}^k$.
We therefore have:
:$\ds \paren {B_{X^\ast} }^k \subseteq \bigcup_{F \in \FF_k} V_{F, k, m}$
By the Banach-Alaoglu Theorem, $\struct {B_{X^\ast}, w^\ast}$ is compact.
Hence from Tychonoff's Theorem, $\struct {\paren {B_{X^\ast} }^k, \paren {w^\ast}^k}$ is compact.
Hence there exists $F_{k, 1}, \ldots, F_{k, N_k} \in \FF_k$ such that:
:$\ds \paren {B_{X^\ast} }^k \subseteq \bigcup_{j \mathop = 1}^{N_k} V_{F_{k, j}, k, m}$
Let:
:$\SS_{k, m} = \set {s_{F_{k, j}, k, m} : 1 \le j \le N_k}$
Hence given $\tuple {g_1, \ldots, g_k} \in \paren {B_{X^\ast} }^k$, there exists $s \in \SS_{k, m}$ such that:
:$\ds \cmod {\map {g_i} {s - x_0} } < \frac 1 m$ for $1 \le i \le k$.
We let:
:$\ds \SS = \bigcup_{\tuple {k, m} \in \N^2} \SS_{k, m}$
From Countable Union of Finite Sets is Countable, $\SS$ is countable.
We now argue that $x_0 \in \map {\cl_w} \SS$.
Let $U$ be a weakly open neighborhood of $x_0$.
From Open Sets in Weak Topology of Topological Vector Space, there exists $\tuple {f_1, \ldots, f_k} \in \paren {X^\ast}^k$ and $\epsilon > 0$ such that:
:$\ds \set {x \in X : \cmod {\map {f_i} {x - x_0} } < \epsilon \text { for each } 1 \le i \le k} \subseteq U$
Then setting:
:$\ds g_i = \frac {f_i} {\norm {f_i}_{X^\ast} }$ for each $1 \le i \le k$
we have $\norm {g_i}_{X^\ast} = 1$ and:
:$\ds \set {x \in X : \cmod {\map {g_i} {x - x_0} } < \epsilon^\ast \text { for each } 1 \le i \le k} \subseteq U$
where we set:
:$\ds \epsilon^\ast = \epsilon \min_{1 \le i \le k} \frac 1 {\norm {f_i}_{X^\ast} }$
Then $\tuple {g_1, \ldots, g_k} \in \paren {B_{X^\ast} }^k$.
Take $m \in \N$ such that $m^{-1} < \epsilon$. 
There exists $s \in \SS$ such that:
:$\ds \cmod {\map {g_i} {s - x_0} } < \epsilon$ for all $1 \le i \le k$.
Hence we have $s \in \SS \cap U$.
Hence $\SS \cap U \ne \O$.
Since $U$ was arbitrary, we have $x_0 \in \map {\cl_w} S$.
{{qed}}