= Solution
The <radial retraction of a punctured ball> gives a <retraction>: project away from the removed interior point until reaching the <unit sphere>. Write $a=x_0$, $v=x-a\ne0$, and set
$$
\tau(x)=\frac{-a\cdot v+\sqrt{(a\cdot v)^2+(1-|a|^2)|v|^2}}{|v|^2},
\qquad
\boxed{r(x)=a+\tau(x)(x-a).}
$$
The equation $|a+tv|^2=1$ is the <quadratic equation>
$$
|v|^2t^2+2(a\cdot v)t+|a|^2-1=0.
$$
Its constant term is negative, so it has exactly one positive root, namely $\tau(x)$. Since $|a+v|=|x|\leq1$, that positive root satisfies $\tau(x)\geq1$. Thus $r(x)\in\partial\mathbb B^n$ for every $x\ne a$ in the <closed unit ball>.
The displayed expression is a <continuous map> on the punctured <closed unit ball>: its denominator is nonzero there, and its <square root> has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a <retraction>. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the <no-retraction theorem>.
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