What's wrong with my proof that $\sigma(a)\subseteq[-\|a\|, \|a\|]$ for $a$ self-adjoint? Let $U$ be a $C^*$-algebra and $a\in U$ be self-adjoint. I have a simple proof that $\sigma(a)\subseteq [-\|a\|,\|a\|]$, where $\sigma(a)$ is the spectrum of $a$.
It goes as follows (the facts used are all known at this point):

We have $\sigma(a) = \sigma(a^*) = \overline{\sigma(a)}$, hence $\sigma(a)\subseteq \mathbb R$. Since $a$ is normal, we have $\|a\| = r(a)$ (where $r(a) = \sup \{|\lambda| : \lambda \in \sigma(a)\})$. Hence $\sigma(a)$ is bounded by $\pm \|a\|$, i.e. $\sigma(a) \subseteq [-\|a\|, \|a\|]$.

The reason why I think this must be wrong is because both Kadison-Ringrose and Bratelli-Robinson use much more elaborate arguments.
What's wrong with my proof?
 A: (1)$\sigma(a)=\overline{\sigma(a)}$ does not imply $\sigma(a)\subset \mathbb{R}$.
To prove this, I'd like use a little functional calculus for Banach algebra.
Since $a$ is hermitian, $e^{ia}=\sum_{n=0}^{\infty}\frac{(ia)^n}{n!}$ is a unitary(note that $(e^{ia})^*=e^{-ia}$ can be examined directly from the formula). 
Hence $e^{i\sigma(a)}=\sigma(e^{ia})\subset \{\lambda:|\lambda|=1\}$. So $\sigma(a)\subset \mathbb{R}$.
(2)If $|\lambda|>\|a\|$, then $\sum_{n=0}^{\infty}\frac{a^n}{\lambda^{n+1}}$ is convergent, and $(\lambda-a)\sum_{n=0}^{\infty}\frac{a^n}{\lambda^{n+1}}=1=\sum_{n=0}^{\infty}\frac{a^n}{\lambda^{n+1}}(\lambda-a)$, hence $\lambda\notin \sigma(a)$. Then we obtain $\sigma(a)\subset [-\|a\|,\|a\|].$
A: I believe the first sentence is wrong. It does imply, that $\lambda \in \sigma(A) \Rightarrow \bar{\lambda}\in \sigma(A)$, but not that in this case $\lambda = \bar \lambda$.
A: Take for instance $\mathbb D$, the unit disk. Then $\overline{\mathbb D}=\mathbb D$ as sets, but that does not make its elements real. 
If for example you take the unilateral shift $S\in B(\ell^2(\mathbb N))$, you have $\sigma(S)=\sigma(S^*)=\text{cl}\,\mathbb D$. 
