Hilbert projections are linear, self-adjoint and contractive

Linearity: for scalars $a,b$ and $x,x^{\mathrm{\prime }}$, the vector $a{P}_{M}x+b{P}_M{x}^{\mathrm{\prime }}$ lies in $M$ and $(ax+b{x}^{\mathrm{\prime }})-(a{P}_{M}x+b{P}_M{x}^{\mathrm{\prime }})=a(x-P_{M}x)+b(x^{\mathrm{\prime }}-P_M{x}^{\mathrm{\prime }})$ lies in $M^{\perp }$, so by the defining property of $P_M$ it equals $P_M(ax+b{x}^{\mathrm{\prime }})$; idempotence follows because $P_{M}x\in M$ has zero orthogonal component, so $P_M{P}_{M}x=P_{M}x$.

Range and kernel: $P_{M}x\in M$ always, $P_{M}m=m$ for $m\in M$ because $m-m=0\in M^{\perp }$, and $P_{M}x=0$ exactly when $x=x-0\in M^{\perp }$; hence $\mathrm{ran}{P}_M=M$ and $\ker P_M=M^{\perp }$.

Self-adjointness: writing $y=P_{M}y+(y-P_{M}y)$ and using additivity in the second argument together with $y-P_{M}y\in M^{\perp }$ and $P_{M}x\in M$ gives $⟨P_{M}x,y⟩=⟨P_{M}x,P_{M}y⟩$, and symmetrically $⟨x,P_{M}y⟩=⟨P_{M}x,P_{M}y⟩$; hence the two pairings agree.

Contractivity: $x=P_{M}x+(x-P_{M}x)$ is a sum of orthogonal vectors, so Pythagoras gives $\mathrm{\parallel }x{\mathrm{\parallel }}^2=\mathrm{\parallel }{P}_{M}x{\mathrm{\parallel }}^2+\mathrm{\parallel }x-P_{M}x{\mathrm{\parallel }}^2\ge \mathrm{\parallel }{P}_{M}x{\mathrm{\parallel }}^2$, hence $\mathrm{\parallel }{P}_{M}x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }{P}_M\mathrm{\parallel }\le 1$ by the definition of the operator norm; if $M\mathrm{\ne }\{0\}$ choose $0\mathrm{\ne }m\in M$, then $P_{M}m=m$ gives $\mathrm{\parallel }{P}_M\mathrm{\parallel }\ge \mathrm{\parallel }{P}_{M}m\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }m\mathrm{\parallel }=1$, so $\mathrm{\parallel }{P}_M\mathrm{\parallel }=1$.