Q:

Banach Spaces with Separable (Continuous) Duals are Separable. A clarification.

I'm aware of this post, but I have not yet been granted permission to comment on these other posts, so I hope you'll excuse asking for a clarification here.
For clarity, I'll restate the question. If $X$ is a Banach space and $X^*$ (the space of continuous linear functionals) is separable, then $X$ is separable. 
In this answer, the chosen answer provides the following argument: Choose a countable dense subset of $X^*$, say $\{f_n\}$. Choose $x_n\in X$ so that $|f_n(x_n)|\ge 1/2||f_n||_{op}$. If $Y$, the rational span of $\{x_n\}$, is not dense, then Hahn-Banach guarantees the existence of a nonzero continuous functional $f$ on $X$ so that $f$ vanishes on $\overline{Y}$. Find an $n$ for which $||f-f_n||_{op}<1/4$, then since $f(x_n)=0$, we have that
$1/4>||f-f_n||_{op}\ge |f(x_n)-f_n(x_n)|=|f_n(x_n)|$
Everything so far is elementary. Now, excuse me if I'm missing something simple, but he/she concludes that $|f_n(x_n)|>1/2$, and immediately reaches the contradiction. But we only know that $|f_n(x_n)|>1/2||f_n||_{op}$, no? 

A:

You can rescale $f$ so that it has unit norm. Then $\|f-f_n\|_{op}<1/4$ implies $\|f_n\|_{op}>3/4$. Now you can finish the chain of inequalities with $|f_n(x_n)|>(1/2)\|f_n\|_{op}>3/8$, which is still a contradiction.

