Let $A$ be a $C^*$-algebra and $a,b\in A$. Therefore $0\leqslant a \leqslant b \Rightarrow \|a\|\leqslant \|b\|$.
I'm trying to prove this claim, but apparently it's necessary to use some spectral theory tools. I got a draft and there's a single step that still isn't cool, but to avoid bias, I will provide a sketch at the end. In Wegge-Olsen's book, it is contained in a single lemma along with the statement that $x\leqslant \|x\|\cdot 1$ for every $x\in A$ (which I was able to prove), in which he simply says "Spectral theory".
I would appreciate any tip. Thanks!
Sketch of my proof. My idea was to use the first statement and conclude that $a\leqslant \|b\|\cdot 1$ and therefore, consider the operator $\lambda \longmapsto \|b\| - \lambda$, and then, using the "positivity" of that operator in $C^*$-world.
Here's a spectral argument. We can reduce to the case where the C$^*$-algebra is unital.
Proof: Being positive, and therefore normal, $\{a,1\}$ generates a commutative unital C$^*$-subalgebra $B$ of $A$. The statement of the lemma is easily verified in $C(X)$ (continuous functions on any compact Hausdorff space $X$), so it holds in $B$ by the Gelfand representation theorem. Since the inclusion $B \hookrightarrow A$ preserves the norm and unit, it follows for $a \in A$. $\Box$
Proof of main result: Since $a \leq b$, for all $\lambda$ such that $b \leq \lambda \cdot 1$, we have $a \leq b \leq \lambda \cdot 1$. Since both are positive, by the lemma, $\| a \| \leq \| b \|$. $\Box$