Here is the offending proof:
$\bf 1.14.$ Proposition. Let $\scr A$ and $\scr B$ be $C^*$-algebras with a common identity and norm such that $\scr A\subseteq \scr B$. If $a\in\scr A$, then $\sigma_{\scr A}(a)=\sigma_{\scr B}(a)$.
$\rm P\scriptstyle{ROOF}$ First assume that $a$ is hermitian and let ${\scr C}=C^* (a)$, the $C^*$-algebra generated by $a$ and $1$. So $\scr C$ is abelian. By Corollary 1.13 $\sigma_{\scr C}(a)\subseteq \Bbb R$. By Theorem VII.5.4, $\sigma_{\scr A}(a)\subseteq\sigma_{\scr C}(a)=\partial\sigma_{\scr C}(a)\subseteq\sigma_{\scr A}(a)$; so $\sigma_{\scr A}(a)=\sigma_{\scr C}(a)\subseteq\Bbb R$. By similar reasoning, $\sigma_{\scr B}(a)=\sigma_{\scr C}(a)$, and hence $\sigma_{\scr A}(a)=\sigma_{\scr B}(a)$.
$\quad$ Now let $a$ be arbitrary. It suffices to show that if $a$ is invertible in $\scr B$, $a$ is invertible in $\scr A$. So suppose there is a $b$ in $\scr B$ such that $ab=ba=1$. Thus, $(a^*a)(bb^*)=(bb^*)(a^* a)=1$. Since $a^* a$ is hermitian, the first part of the proof implies $a^* a$ is invertible in $\scr A$. But inverses are unique, so $bb^*=(a^* a)^{-1}\in\scr A$. Hence $b=b(b^*a^*)=(bb^*)a^*\in\scr A$.
In that inclusion chain, in the middle, we have:
$$\sigma_{\mathcal{C}}(a)=\partial\sigma_{\mathcal{C}}(a).$$
The question obviously is: what does $\partial$ mean? To me, it means the boundary, which, in my book, is the set of points which are limit points for both the set and the complement. So this equality is stating that all points of $\sigma_{\mathcal{C}}(a)$ are limit points both for that spectrum and its complement. And then the question is: how is that proven? I do know that, since $a$ is hermitian in that part, the spectrum is real. But that only guarantees that all its points are limit points for its complement. Of course, I might have misinterpreted $\partial$. So why is this equality true? Is there a proof (maybe somewhere in Conway), or is it just that $\partial$ indicates the limit points of the complement, and not what I call boundary and denote with $\partial$?
$\def\c{\complement} \def\cl{\mathrm{cl}\,}$ Taking this definition: $\partial A:=\cl A\cap \cl A^\c$, the spectrum $A:=\sigma(a)$ is closed, and its complement is dense as misses at most parts of the real line.
This proves $\sigma(a)=\partial\sigma(a)$.