Let $X$ be a separable Banach space, and $(f_n)$ be a countable dense subset. Recall that for each $f_n$ there exists a linear functional $l_n \in X^*$ such that $\|l_n\| = 1$ and $l_n(f_n)=\|f_n\|$. Now define a new linear operator $g: X \to L^\infty(\mathbb N)$ as follows. $$g(f) := \{l_n(f)\}_n.$$ Show that $$\|g(f)\| := \sup_n \left|l_n(f)\right| = \|f\|.$$ That is, $g$ is an isometry and hence injective.
I know that $$\|f\| = \sup_{l \in X^*,\ l\neq 0} \frac{\left|l(f)\right|}{\|l\|} \geq \sup_n \left|l_n(f)\right|=:\|g(f)\|.$$ It is only left to show the opposite direction. How to proceed from here, please? Thank you!
Write $f=\lim_{k\rightarrow\infty}f_{n_{k}}$ since $\left\{f_{n}\right\}$ is dense in $X$. Then $$ \left\Vert g\left(f\right)\right\Vert =\sup_{n}\left|l_{n}\left(f\right)\right|=\sup_{n}\left|l_{n}\left(\lim_{k\rightarrow\infty}f_{n_{k}}\right)\right|. $$ Since $l_{n}$ is a bounded linear functional, $l_{n}$ is continuous and hence $$ \left\Vert g\left(f\right)\right\Vert =\sup_{n}\left|\lim_{k\rightarrow\infty}l_{n}\left(f_{n_{k}}\right)\right|=\sup_{n}\lim_{k\rightarrow\infty}\left|l_{n}\left(f_{n_{k}}\right)\right|. $$ Since $\left\Vert l_{n}\right\Vert =1$, $$ \left\Vert g\left(f\right)\right\Vert \leq\sup_{n}\lim_{k\rightarrow\infty}\left\Vert f_{n_{k}}\right\Vert . $$ Since $\left\Vert \cdot\right\Vert $ is continuous, $$ \left\Vert g\left(f\right)\right\Vert \leq\sup_{n}\left\Vert \lim_{k\rightarrow\infty}f_{n_{k}}\right\Vert =\sup_{n}\left\Vert f\right\Vert =\left\Vert f\right\Vert . $$