Let $c_0(\mathbb N)$ be the space of sequence in $\mathbb C$ whose limit is zero, equipped with the $\ell^\infty$ norm. Let $u_n$ be a sequence in $\mathbb C$ and define the operator $A$ taking a sequence $x_n$ in $\ell ^\infty (\mathbb N)$ to the sequence $u_nx_n$. I need to prove $A\in B(\ell^\infty(\mathbb N), c_0(\mathbb N)$ iff it's in $K(\ell^\infty(\mathbb N), c_0(\mathbb N)$.
$\impliedby$ always holds, and for $\implies$, I think the strategy is the same as here, but again, I'm having trouble showing why boundedness is needed for compactness. So letting $P_n$ be the operator that truncates after the $(n+1)^\text{th}$ place, I want to show $\|P_nAP_n-A\|\rightarrow 0$. Now
$$\|P_nAPx-Ax\|_\infty=\ \sup_{\|x\|=1}\|(0,\dots,0,u_{n+1}x_{n+1},\dots)\|_\infty $$ so if $u_nx_n\rightarrow 0$ I think that alone suffices to show the RHS tends to zero, and in fact the convergence implies boundedness. Finally, $P_nAP_n$ is compact because it has finite rank, so $A$ is compact as a norm-limit of compact operators.
Did I actually need $A$ to be bounded?