Compact operators, space of sequences Let $\phi\in\ell^\infty$. For $p\in[1,\infty]$, define $M_\phi:\ell^p\to\ell^p$ by
$$M_\phi(f)=\phi f.$$
Show that $\Vert M_\phi\Vert=\Vert\phi\Vert_\infty$, and $M_\phi$ is compact if and only if $\phi\in c_0$, i.e. $\phi$ is a sequence that converges to $0$.
I only have problem with the part "$\phi\in c_0$ $\Rightarrow$ $M_\phi$ compact.
I tried to prove by contradiction, assume $M_\phi$ is not compact, then there is a bounded sequence $(f_n)_{n\in\mathbb{N}}$ in $\ell^p$ s.t. $(\phi f_n)_{n\in\mathbb{N}}$ has no convergent subsequence, then it also has no Cauchy subsequence. Then we can define
$$t:=\inf_{m\neq n}\Vert \phi(f_m-f_n)\Vert_p>0$$
At this point I am stuck, I tried to find some $n\neq m$ s.t. $\Vert \phi(f_m-f_n)\Vert_p<t$, then we get a contradiction. Can someone give some hints? Thanks!
P.S. It is required to use the definition of compactness to prove this question.

If $\boldsymbol\varphi\in\ell^\infty(\mathbb N)\smallsetminus c_0(\mathbb N)$, then there is an $\varepsilon>0$, and a subsequence $\{\varphi(k_n)\}_{n\in\mathbb N}$, such that
$$
\lvert \varphi(k_n)\rvert\ge \varepsilon.
$$
Let now 
$$
\boldsymbol{u}_n=\boldsymbol{e}_{k_n},
$$
where $\boldsymbol{e}_{n}$ is the sequence is which is zero for all $k\in\mathbb N$, except $k=n$, where it takes the value 1. Clearly, $\boldsymbol{e}_{n}\in\ell^p(\mathbb N)$, for all $p$, and $\|\boldsymbol{e}_{n}\|_p=1$. Now $\{\boldsymbol\varphi\boldsymbol{u}_n\}_{n\in\mathbb N}$ does not have a convergent subsequence as
$$
\|\boldsymbol\varphi\boldsymbol{u}_m-\boldsymbol\varphi\boldsymbol{u}_n\|_p\ge 2^{1/p}\varepsilon,
$$
for all $m\ne n$.
