= Solution
Let $Z=\overline{\operatorname{span}}\{x_n:n\geq1\}$. If $Z\ne X$, part (a) provides a nonzero bounded functional annihilating $Z$; normalize it to have <norm> one. Density of the sequence in the dual unit ball gives an index $n$ with $\|f_n-f\|<1/4$. Since $f(x_n)=0$ and $\|x_n\|\leq1$,
$$
|f_n(x_n)|\leq\|f_n-f\|<\frac14.
$$
But the hypothesis and reverse <triangle inequality> give
$$
|f_n(x_n)|\geq\frac12\|f_n\|>\frac12\left(1-\frac14\right)=\frac38,
$$
a contradiction. Therefore \b[the span is dense].
If $X'$ is <separable>, its unit ball has a countable <dense set>. For each nonzero $f_n$, the definition of its <norm> permits choosing $x_n$ in the unit ball with $|f_n(x_n)|\geq\|f_n\|/2$; choose zero for a zero functional. The preceding result gives a dense countable span over $\mathbb Q$, or over $\mathbb Q+i\mathbb Q$ in the complex case. Thus \b[$X$ is <separable>].
The converse fails: $\ell^1$ is <separable>, since finite rational-coordinate sequences are dense, but $(\ell^1)'\cong\ell^\infty$ is not <separable>. The dual identification sends a bounded sequence $a$ to $x\mapsto\sum_na_nx_n$, isometrically; conversely a bounded functional is determined by its values on the <coordinate vectors> and hence has this form by truncation. The uncountable family of zero-one sequences in $\ell^\infty$ has pairwise <distance> one, which no countable <dense set> can approximate within a radius smaller than $1/2$.
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