$(B_X,w)$ metrizable implies $X^\ast$ separable Let X be a normed space and assume that $(B_X,w)$ is metrizable, i.e. the weak topology is metrizable. Show that $X^\ast$ is separable. 
My attempt:
Let $d$ a equivalent metric on $B_X$. For fixed $N\in\mathbb{N}$ there exists a finite family of functionals $F_N\subset X^\ast$ such that $U_N:=\{ x \in X: d(x,0) < N^{-1} \} \supset \{x\in X: |f(x)|<\epsilon_N ,f\in F_N\}$. Define $Z:= \overline{\text{span}}\{ F_N: N\in\mathbb{N}\}$. If $Z=X^*$, we are done; just take the rational span. Now assume that $g\in X^\ast$. Then there exists a $n\in \mathbb{N}$ such that $U_n \subset \{x: |g(x)|< 1\}$. If $x\in \ker f$ for all $f\in F_n$, then $\lambda x \in U_n$ for all $\lambda\in \mathbb R$. Then $|g(x)|<\frac{1}{|\lambda|}$ for all $\lambda \in \mathbb{R}$ and hence, $x\in \ker g$. As $\cap_{f\in F_n} \ker f \subset \ker g$, $g\in \text{span} F_n$ and thus $g\in Z$.
Is the last argument okay. In particular, the argument with the $\lambda$ and the kernel.
Edit:
Claim: If $\ker g\supset \cap_{i=1}^n \ker f_i =: Y$ for linear functionals in $X^*$. Then, $g\in \text{span} \{f_1, \cdots f_n\}$.
Proof. Consider the map  $\vec{f}: X\to \mathbb{R}^n, x \mapsto (f_1(x),\dots, f_n(x))$, i.e. $Y=\ker \vec{f}$. As $Y\subset \ker g$, there exists a unique map $\hat g: \mathbb{R}^n \to \mathbb{R}$ such that  $g=\hat g \circ \vec f$. Note $\mathbb{R}^n = X/Y$. Now, $g(x)=\sum_{i=1}^n \lambda_i x_i$ for $x\in\mathbb{R}^n$ and some fixed $\lambda_i\in \mathbb{R}$. Thus, $g(x)=\sum_{i=1}^n \lambda_i f_i(x)$, which proves the claim.
Edit 2: This proof is definitely false as it contradicts the Baire category theory. We had a countable Hamel basis which cannot be true. See the comment.
Edit 3: The Problem is resolved. I assumed that $X$ and not only the ball $B_X$ was metrizable. Thanks to Nik Weaver who pointet that out on MO. 
 A: A few comments: 
1) you don't need "contradiction". Your argument (if right) shows that if $g\in X^*$, then $g$ is in the span of $F_N$. 
2) I don't see why you claim that $\cap_{f\in F_N}\ker f\subset\ker g$ implies $g\in \text{span}\,F_N$.
3) You seem to prove that $X=\text{span}\,\{F_N:\ N\}$ without closure, which looks suspicious to me. 
4) Here is a way to fix the proof: 
If $g\in X^*\setminus Z$, then we can separate them by Hahn-Banach (considering the locally convex space $X^*$ with the weak$^*$-topology), so there exists $x\in X$ and $c\in\mathbb R$ with
$$\tag{1}
\text{Re}\,f(x)<c\leq\text{Re}\,g(x).
$$
As $Z$ is a subspace, it follows that $\text{Re}\,f(x)=0$ for all $f$ in $Z$, and in particular $-\text{Im}\,if(x)=\text{Re}\,f(x)=0$, so $f(x)=0$ for all $f\in Z$. 
Now we can argue as in your attempt. There exists $n$ such that $U_n\subset\{y:\ |g(y)|<\|x\|+1\}$. As $\lambda x\in U_n$ for all $\lambda\in\mathbb R$, we get 
$$
|g(x)|\leq\frac{\|x\|+1}\lambda,\ \ \lambda\in\mathbb R,
$$
implying that $g(x)=0$, contradicting (1). So $g$ cannot exist, and $X^*=Z$.
