Consider the following theorem in the book "$C^*$-algebras and operator theory" written by Murphy.
Questions: (1) Is the theorem talking about orthogonal projections? (as defined in the text above)? Or simply projections?
(2) How to prove $(2) \implies (6)?$ I assume that the theorem talks about orthogonal projections, and then I have difficulties. I can prove that $(q-p)^2 = q-p$ but I do not succeed in proving that $q-p$ is orthogonal (when $p,q$ are).
On
Here "projection" refers only to orthogonal projections. This is a very common convention in the context of operators on Hilbert spaces.
To prove (2) implies (6), note that an operator $p$ is a projection (meaning orthogonal projection) iff $p^2=p^*=p$. You have verified that $(q-p)^2=q-p$, and $q-p$ is trivially self-adjoint since $p$ and $q$ are.
(To prove this characterization of orthogonal projections, note that if $p^2=p^*=p$ then $\langle pv,(1-p)w\rangle=\langle v,p^*(1-p)w\rangle=\langle v,p(1-p)w\rangle=0$ for all $v,w\in H$ and thus the images of $p$ and $1-p$ are orthogonal subspaces of $H$ whose sum is all of $H$ (since $v=pv+(1-p)v$), with $p$ being the projection onto the image of $p$ with respect to this splitting.)
The theorem refers about orthogonal projections. In Operator Algebras it is standard to reserve "projection" for orthogonal projection, and "idempotent" when one wants to talk in general.
The key result needed is that an idempotent is orthogonal if and only if it is selfadjoint. This is trivial: if $p$ is an orthogonal projection, then we have $px\perp (1-p)x$ for all $x$; this means that, for all $x,y$, $$ 0=\langle px,(1-p)y\rangle=\langle (1-p)^*px,y\rangle. $$ Thus $0=(1-p)^*p=(1-p^*)p=p-p^*p$. Then $p=p^*p$ is selfadjoint. Conversely, if $p$ is a selfadjoint idempotent, then $(1-p)p=0$ by the above computation, and thus $px\perp (1-p)y$ for all $x,y$ and $p$ is orthogonal.
So, if $p,q$ are orthogonal projections with $pq=q$, then $p-q$ is a selfadjoint idempotent and thus an orthogonal projection.