 Solution

The vector subspace contains zero, whose approximation error is $\Vert u\Vert$. For any $q$ with $\Vert q\Vert >2\Vert u\Vert$, the triangle inequality gives $\Vert u-q\Vert \ge \Vert q\Vert -\Vert u\Vert >\Vert u\Vert$, so such points cannot improve on zero.

Restrict to $K=\{q\in A:\Vert q\Vert \le 2\Vert u\Vert +1\}$. This is nonempty, closed and bounded in a finite-dimensional normed space, hence compact. The continuous function $q\mapsto \Vert u-q\Vert$ attains a minimum on $K$ by part (a). Outside $K$ its value exceeds the error at zero, so that minimizer is best on all of $A$:Completeness of the ambient space is unnecessary; compactness is used only in the finite-dimensional vector subspace.