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I'm aware that this question has been answered here Every separable Banach space is isometrically isomorphic to a subspace of $\ell_\infty$

But I want to prove this result a different way.

Suppose that $\mathcal{B}$ is separable and define $i : \mathcal{B} \to L^{\infty}(\mathbb{N})$ in the following way: For a dense subset of the unit ball in $\mathcal{B}$, denoted by $\{ f_n \}$, define $x_n = (\cdot, f_n)$ and thus define $i(y) = ((y,f_n))_{n=1}^{\infty}$. It is clear that this map is linear, injective and that $\| i(y) \|_{L^{\infty}(\mathbb{N})} = \| y \|_{\mathcal{B}}$. It remains to show that $i$ is bounded. This is where I'm stuck.

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You are on the way. Anyway you have to choose properly $f_n$. There is a dense subset $D = x_n$ Now using hahn banach you can find $f_n$ such that $f(x_n)=\Vert x_n \Vert $ and the norm of $f_n$ is 1. Take $i(y) = (f_n(y))_n$. We know that $\Vert y \Vert = sup_{\Vert f \Vert \le 1} f(y)$. Using densness of $x_n$ You get that $i$ is an isometry, and you finish

Edit More correctly. Fix $y$. We know that there exists $x_{n_k}$ which converges to $y$. Consider $f_{n_k}$. From the metrizability of the unit ball in the weak-* we know that there is a subsequence that converges weakly-*. Thus $f_{n_k}(x_{n_k}) \to f(y)$, for some f with $\Vert f \Vert \le 1$. But $f_{n_k}(x_{n_k}) =\Vert x_{n_k} \Vert \to \Vert y \Vert$

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