$c$ and $c_0$ are not isometrically isomorphic

The idea is that a linear isometry $j: V\rightarrow W$ is injective and sends extreme points of any convex subset $K\subset V$ to extreme points of $j(K)\subset W$,
in fact $j(K)$ is convex, for $j(u), j(v)\in j(K): tj(u)+(1-t)j(v)= j(tu+(1-t)v)\in j(K)$

Recall the definition of extreme points : $x$ is extreme in $C$ when $\exists\ a,b\in C \ x= ta+(1-t)b \implies a=b= x$

Then if $x\in K$ is extreme $$j(x)= tj(u)+(1-t)j(v)\implies j(x)=j(tu+(1-t)v) \implies x= tu+(1-t)v \implies u=v=x \implies j(u)=j(v)=j(x)$$ So $j(x)$ is extreme in $j(K)$ , and conversely as well if $j(y)$ is extreme then so is $y$ .

You can now apply this to $c_0$ and $c$, recall that every normed linear space is locally convex. In particular, every closed ball is convex.

The unit closed ball of $c_0$ has no extreme points, if $(x_n)$ is in this ball then $\exists N\in \mathbb{N}: \forall n\geq N\ |x_n|<1/5$ and we can construct $(y_n)$ and $(z_n)$ to be in this ball with $(x_n)= 1/2\left((y_n)+ (z_n)\right)$ by simply 'perturbing' just the $N^{th}$ term.

While the unit closed ball of $c$ has $(a_n)= (1,1,1,1,\dots)$ as an extreme point (why?) .

So having a different number of extreme points, what can you conclude? (be careful if you need your map to be surjective)