This is almost the same as the other answers, with a bit of detail:
Let X be a separable Banach space.
Let $(x_n)$ be a dense sequence in the unit sphere of $X$.
For each $n$, use the Hahn-Banach Theorem to find a norm-one functional $f_n\in X^*$ with $f_n^*(x_n)=1$.
Define $\Phi: X\rightarrow\ell_\infty$
via $$
\Phi(x) =(f_n(x) )
$$
$\Phi$ is clearly linear.
Suppose $x\in X$ has norm one and let $1>\epsilon>0$. Choose $n_\epsilon$ so that $\Vert x_{n_\epsilon}-x\Vert<\epsilon$.
Then
$$\epsilon>|f_{n_\epsilon}(x_{n_\epsilon}-x)|=|f_{n_\epsilon}(x_{n_\epsilon})-f_{n_\epsilon}(x)| = |1-f_{n_\epsilon}(x)|.$$
As $\epsilon$ was arbitrary, this implies that $\Vert \Phi(x)\Vert=\sup\limits_{n\in \Bbb N}|f_n(x ) |\ge 1$.
Also,
for any $n$, $$|f_n(x)|\le \Vert f_n \Vert \Vert x\Vert =1.$$
and so $\Vert \Phi(x)\Vert\le 1$.
Thus we have $\Vert \Phi(x)\Vert= 1$, whenever $\Vert x\Vert =1$.
From this it follows that for any non-zero element $x$ of $X$, we have
$$\Vert \Phi(x)\Vert= \Vert x\Vert \sup\limits_{n\in \Bbb N}|f_n(x/\Vert x\Vert) | =\Vert x\Vert;$$ and thus, $\Phi$ is an isometry.