The diameter of the set of projections in a C*-algebra is at most 1

Question

Let $p,q$ be projections in a $C^*$-algebra $A$. I am trying to show that $\|p-q\|\leq1$, but I can't.

If the projections $p,q$ commute, then this is easy: we set $C=C^*(1,p,q)$ and this is an abelian $C^*$-algebra. By the Gelfand representation, we have that $\sigma(x+y)\subset\sigma(x)+\sigma(y)$ in an abelian C*-algebra, thus $\sigma_A(p-q)=\sigma_C(p-q)\subset\sigma_C(p)-\sigma_C(q)\subset\{-1,0,1\}$ and therefore $\|p-q\|\leq1$. But what about the general case?

Answer 1

For projections $p,q \in A$ we have $0 \le p,q \le 1$ so

$$-1 \le -q \le p-q \le p \le 1$$

which implies $\sigma(p-q) \subseteq [-1,1]$ and therefore $\|p-q\| \le 1$.