does this theorem imply that every $C^*$-algebra has an approximate unit?

Question

I read that every $C^*$-algebra has a approximate unit. But we proved only the following theorem in lecture:

Let $A$ be a $C^*$-algebra, $I\subseteq A$ an ideal which is dense in $A$. Then there is an approximate unit $(u_\lambda)_{\lambda\in\Lambda}$ in $A$ such that :
(i)$0\le u_\lambda$, $u_\lambda\in I$ and $\|u_\lambda\|\le 1$ for every $\lambda\in \Lambda$
(ii)if $\mu\le \lambda$, then $u_\mu\le u_\lambda$

My question is: Does this theorem imply that every $C^*$-algebra has an approximate unit?
What I think about it: The question reduces to: does every $C^*$-algebra $A$ have an ideal $I$ which is dense in $A$? The answer is yes, we could take $I=A$. Therefore the the theorem implies that every $C^*$-algebra has an approximate unit.

Is my argument correct? I'm sceptical, because the argumentation seems to be too easy.

Answer 1

My understanding is that first you show that every C*-algebra has an approximate unit and use this to prove your result here.

You do this by working with the C*-algebra $B=I^*\cap I$... so your argument is circular, yes.