Here is a simple proof that the
Banach-Mazur distance between
the spaces $c$ and $c_0$
is equal to $3$.

First the map $T:c \to c_0$
defined by
\begin{align*}
T(x)=\left(3x(\omega) , \frac{3}{2}\big(x(1)- x(\omega)\big), \frac{3}{2}\big(x(2)- x(\omega)\big),\ldots \right)
\end{align*}
is an isomorphism with $\|T\|\|T^{-1}\|\le 3$.

To show that we can't do better, suppose there is an isomorphism $T:c \to c_0$ with $\|T^{-1}\|\le 1$ and $\|T\|\le K<3$. Let $e_0=(1,1,1,\ldots)$, and $(e_i)_{i\ge 1}$ be the standard unit vectors. Let $\varepsilon=\frac{3-K}{2}$. Let $N$ be such that $|Te_0(t)|<\varepsilon$ for all $t>N$. By pigeonhole principle there exists (in fact, infinitely many) $i_0$ such that $|T(e_{i_0})(t)|<\varepsilon/2$ for all $t\le N$. Consider the vector $T(e_0+2e_{i_0})$. Since $\|e_0+2e_{i_0}\|=3$ we have $\|T(e_0+2e_{i_0})\|\ge 3$. The norm must be 'attained' somewhere so let's check if it is attained at some $t\le N$.
$$3\le |T(e_0+2e_{i_0})(t)|\le |T(e_0-e_{i_0})(t)|+3|T(e_{i_0}(t))|<|T(e_0-e_{i_0})(t)|+3\varepsilon/2$$
So $|T(e_0-e_{i_0})(t)|>3-3\varepsilon/2$. But this is impossible since $|T(e_0-e_{i_0})(t)|\le K$ (check that $K>3-\frac{3\varepsilon}{2}$ not possible for $K<3$).

On the other hand, if the norm is attained at some $t>N$ we have
$$3\le |T(e_0+2e_{i_0})(t)|<\varepsilon +2|T(e_{i_0})(t)|$$ so
$$|T(e_{i_0})(t)|>\frac{3-\varepsilon}{2}$$
But since $\|e_0-2e_{i_0}\|=1$ we have
$$K\ge \|T(e_0-2e_{i_0})\|\ge 2|T(e_{i_0})(t)-|T(e_0(t))|\ge 2\frac{3-\varepsilon}{2}-\varepsilon=3-2\varepsilon$$
again leads to a contradiction

isisometrically isomorphic to $c$ as Banach spaces, but the isomorphism does not preserve the $C^*$-algebra structures. Also, one choice for your functional $\phi$ is given by $\phi(x_n)=\lim_{n\to\infty} x_n$. This does not "correspond to a bounded sequence $(t_n)$." $\endgroup$ – Teri May 13 '18 at 11:51isometricallyisomorphic. $\endgroup$ – Teri May 13 '18 at 12:35