This is from Analytic K-Homology by Higson and Roe.
Show that if $S$ is a rank one operator and $T$ is any operator, then $ST$ and $TS$ are rank one operators. Deduce that $\mathcal{K(H)}$ forms a closed ideal in $\mathcal{B(H)}$.
From link, link these posts, it seems that my proof given below is too simple to be correct.
Proof: Consider $K\in \mathcal{K(H)},\ T\in \mathcal{B(H)}$. $\exists$ finite rank operators $K_n$ such that $\| K_n-K\| \rightarrow 0$. $\implies \| K_n-K\|\|T\|\rightarrow 0$. $\implies \| K_nT-KT\|\rightarrow 0$. So $KT$ is the norm limit of the finite rank operators $K_nT$, hence compact.
Any help is appreciated.