= Solution
The <retraction of a Hilbert ball onto its sphere> exists. In the real <Hilbert space> $\ell^2$, let $S(x_0,x_1,\ldots)=(0,x_0,x_1,\ldots)$ be the <unilateral shift operator>, and let $e_0=(1,0,\ldots)$. For $x$ in the <closed unit ball> $\mathbb B$, define
$$
F(x)=\sqrt{1-\|x\|^2}\,e_0+Sx.
$$
The <orthogonality> of $e_0$ and $Sx$ gives $\|F(x)\|^2=1$. Thus $F$ is a <continuous map> into the <unit sphere>. It has no <fixed point>: $F(x)=x$ would imply $x_{j+1}=x_j$ for every $j\geq0$. A constant <sequence> in $\ell^2$ must be zero, whereas $F(0)=e_0\ne0$.
Put $d=x-F(x)$, which is nonzero everywhere, and define
$$
\alpha(x)=\frac{2(1-\langle F(x),x\rangle)}{\|x-F(x)\|^2},
\qquad
\boxed{r(x)=F(x)+\alpha(x)(x-F(x)).}
$$
This is a <continuous map> on the whole <closed unit ball>, including zero. To check its range, observe that $\langle F(x),d\rangle=\langle F(x),x\rangle-1$, so
$$
\|F(x)+\alpha d\|^2
=1+2\alpha(\langle F(x),x\rangle-1)+\alpha^2\|d\|^2=1.
$$
If $\|x\|=1$, then $\|x-F(x)\|^2=2(1-\langle F(x),x\rangle)$, hence $\alpha(x)=1$ and $r(x)=x$. Therefore $r:\mathbb B\to\partial\mathbb B$ is a <retraction>.
Geometrically this is the second intersection of the line from $F(x)$ through $x$ with the <unit sphere>. In fact
$$
\alpha(x)-1=\frac{1-\|x\|^2}{\|x-F(x)\|^2}\geq0.
$$
The construction illustrates why the finite-dimensional hypothesis in the <no-retraction theorem> matters. The <closed unit ball> of $\ell^2$ is not a <compact set> in the <norm topology>.
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