Exercise: Let $H$ be an inner product space and $\emptyset \neq S \subset H$. Given $x \in H$ and $y \in S$, show that if $x - y \perp S$, then $$||x-y|| \leq ||x-z||, \ \ \forall z \in S.$$
Attempt: Take $x \in H$ and $y \in S$ such that $x-y \perp S$. Let $z \in Z$ be arbitrary. Then, \begin{align*} ||x-y||^2 & = |\langle x-y, x-y\rangle|\\ & = |\langle x-y,x\rangle - \langle x-y,y\rangle|\\ & = |\langle x-y,x \rangle - \langle x-y,z\rangle|\\ & = |\langle x-y,x-z\rangle|\\ & \leq ||x-y||||x-z||. \end{align*}Now notice that if $||x-y|| = 0$ ,then the inequality is satisfied trivially. So, assuming $||x-y|| \neq 0$, we can divide both sides and we get the result.
Question: Can anyone check my proof?