Radial projection onto the closed unit ball

Question

Let $(E, |\cdot|)$ be a normed vector space. We define the radial projection onto the closed unit ball $T:E \to E$ by $T(x)=x$ if $|x| \le 1$ and $x/|x|$ if $|x|>1$. I'm trying to solve below exercise, i.e.,

1. Prove that $|T(x)-T(y)| \le 2 |x-y|$ for all $x,y\in E$.
2. What happens if $|\cdot|$ is induced from an inner product $\langle \cdot, \cdot \rangle$?

Could you verify my below attempt?

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1. Let $x, y \in E$. The statement is clear if $|x|, |y|\le 1$. First, we assume $|x|, |y| > 1$. Then $$ \begin{align} |T(x)-T(y)| &= \left | \frac{x}{|x|} - \frac{y}{|y|} \right | \\ &= \left | \frac{x|y| - y |x|}{|x| |y|} \right | \\ &= \left | \frac{x(|y| - |x|) + (x - y) |x|}{|x| |y|} \right | \\ &\le \frac{2|x - y|}{|y|} \le 2|x - y|. \end{align} $$

Second, we assume $|x| \le 1$ and $|y| > 1$. Then $$ \begin{align} |T(x)-T(y)| &= \left | x - \frac{y}{|y|} \right | \\ &= \left | \frac{x|y| - y }{|y|} \right | \\ &\le \frac{|x| (|y|-1)+ |x - y| }{|y|} \\ &\le \frac{|x|(|y|-|x|)+ |x - y| }{|y|} \\ &\le \frac{(|x|+1)(|y-x|)}{|y|} \\ &\le 2|y-x|. \end{align} $$

2. Let's prove that $|x-T(x)| \le |x-y|$ for all $y\in E$ with $|y| \le 1$. It is equivalent to prove that $\langle x-T(x), y-T(x) \rangle \le 0$ for all $y\in E$ with $|y| \le 1$. WLOG, we assume $|x| > 1$. Then $T(x) = x / |x|$. Indeed, $$ \begin{align} \langle x-T(x), y-T(x) \rangle &\le 0 \\ \iff \langle x-x/|x|, y-x/|x| \rangle &\le 0 \\ \iff \langle x, y-x/|x| \rangle &\le 0 \\ \iff \langle x, y \rangle \le |x|. \end{align} $$

The last inequality is true by Cauchy–Schwarz inequality and the fact that $|y| \le 1$.