Let $A$ be a non-unital C*-algebra and let $\pi : A \to \mathcal{B}(H)$ be a non-degenerate representation of $A$ (that is, $\mathrm{ span }\{\pi(a)h : a \in A, h \in H\}$ is a dense subset of $H$). Now let $(u_\lambda)_{\lambda \in \Lambda}$ be an approximate unit for $A$. A standard argument, using non-degeneracy, shows that $(\pi(u_\lambda))_{\lambda \in \Lambda}$ converges strongly to $1_H$, the identity operator on $H$. There are several examples that show that $(\pi(u_\lambda))_{\lambda \in \Lambda}$ is not norm convergent.
However, below I will present an argument that "shows" that $(\pi(u_\lambda))_{\lambda \in \Lambda}$ converges in norm to $1_H$. This is certainly at fault, but I want to make sure I understand exactly where the proof goes wrong.
Argument: Let $\epsilon >0$ and choose $h \in H$ with $\|h\|=1$ such that $$ \| 1_H - \pi(u_\lambda) \| < \epsilon + \| h - \pi(u_\lambda)h \|. $$ Then, since strong convergence of $(\pi(u_\lambda))_{\lambda \in \Lambda}$ implies that $(\| h - \pi(u_\lambda)h \|)_{\lambda \in \Lambda}$ converges to $0$, we find $\lambda' \in \Lambda$ such that for all $\lambda \geq \lambda'$, $$ \| h - \pi(u_\lambda)h \| < \epsilon. $$ Thus, for any $\lambda \geq \lambda'$, we have shown $$ \| 1_H - \pi(u_\lambda) \| < 2\epsilon, $$ which proves that $\pi(u_\lambda)$ converges in norm to $1_H$. End of argument.
I am pretty sure that the problem in the above argument is that $h$ depends on $\epsilon$ and therefore also $\lambda'$. Thus, the expression $\| h - \pi(u_\lambda)h \| < \epsilon$ for $\lambda \geq \lambda'$ doesn't make much sense to me. In any case, I am still not quite convinced about this and I would love to hear someone else's thoughts.
Thanks in advance!
Yes, it is as you say. Something that often helps when in doubt with an argument like this is to consider a concrete example.
Let $A=c_0$. Then an approximate unit is given by $u_n=\sum_{k=1}^n e_k$, where $\{e_k\}$ is the canonical basis. We can think $c_0\subset\ell^\infty(\mathbb N)$ (or if you insist on representing on a Hilbert space you can represent all this as diagonal operators).
Now we see what happens. We have $\|1-u_n\|=1$ for all $n$. The $h$ you choose can be for instance $h=e_{n+1}$; there are many other possible choices, but they will all depend on $n$. Now you have $$ \|1-u_n\|=\|e_{n+1}-u_ne_{n+1}\| $$ (that is, we can even take $\epsilon=0$). And you can see that taking limit on the right does not work as your argument says, because the $h=e_{n+1}$ changing with $n$ does not make the norm go to zero.