Given any $r\in{}\mathbb{Q}$, let $r^*$ denote the Dedekind cut $r^*=\{q|q<r\}$.

Claim: For $r\in{}\mathbb{Q}^+$ and $s\in{}\mathbb{Q}^-$, $r^*\cdot{}s^*\subset{}(rs)^*$

Attempt at Proof: $q\in{}r^*\cdot{}s^*\implies{}q\in{}-[r^*\cdot{}(-s)^*]\implies{}\forall{}r'\forall{}s'(r',s'\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,r'\in{}r^*\,\,\,\wedge{}\,\,\,s'\in{}(-s)^*\implies{}\exists{}a(a\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,-q-a>r's'))$

At this point I am stuck. I need to show $rs>q$ so that $q\in{}(rs)^*$. However, it's not valid to claim $rs>-(r's')>q+a$ using the attempt above.

This post covers the question, but I think the proof is wrong at the relevant place and I'm stuck.

From Walter Rudin's Principles of Mathematical Analysis, Third Edition, page 20, Step 8:

I want to complete the proof of (b) by proving the following claim.

Claim: For $r\in{}\mathbb{Q}^+$ and $s\in{}\mathbb{Q}^-$, $r^*\cdot{}s^*\subset{}(rs)^*$

Attempt at Proof: $q\in{}r^*\cdot{}s^*\implies{}q\in{}-[r^*\cdot{}(-s)^*]\implies{}\forall{}r'\forall{}s'(r',s'\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,r'\in{}r^*\,\,\,\wedge{}\,\,\,s'\in{}(-s)^*\implies{}\exists{}a(a\in{}\mathbb{Q}^+\,\,\,\wedge{}\,\,\,-q-a>r's'))$

At this point I am stuck. I need to show $rs>q$ so that $q\in{}(rs)^*$.

This post covers the question, but I think the proof is wrong at the relevant place and I'm stuck.