If $A$ is a C$^{*}$-algebra and $S\subset A$, prove that $ASA$ is a closed ideal

Question

The question is in the title. It is mentioned passively in Remark 5.4.2. in Murphy's C$^{*}$-algebra book. Clearly $ASA$ is an ideal in $A$, but I don't see immediately why it is closed.

Answer 1

It is not. Example: $A=B(H)$, $S$ the finite-rank operators.

But, in Murphy's remark, you can take $J_1=\overline{AS_1A}$, and the argument goes through.