= Solution
{c}
The geometry of a strictly convex <normed space> proves <uniqueness> whenever a <best approximation> exists. If distinct $p_1,p_2$ both have best error $E>0$, then their two residuals have <norm> $E$. The defining midpoint inequality gives
$$
\left\|u-\frac{p_1+p_2}{2}\right\|
=\left\|\frac{(u-p_1)+(u-p_2)}2\right\|<E,
$$
contradicting <best approximation> because the midpoint lies in $A$. If $E=0$, both approximants equal $u$.
The printed <existence> assertion is false for an arbitrary <vector subspace>. For a concrete counterexample take $B=\ell^2$ with its strictly convex Hilbert <norm>, let $A=c_{00}$ be its finite-support sequences, and take $u=(2^{-j})_{j\geq1}$. Truncations in $A$ tend to $u$, so the best-error <infimum> is zero, but $u\notin A$ and that <infimum> is not attained. If the finite-dimensional assumption of part (b) is retained, it supplies <existence> and the argument above supplies <uniqueness>.
For the estimate, suppose $Y$ is a <projection operator> onto $A$, as needed for a finite <operator norm>. Since $Yp=p$,
$$
u-Yu=(u-p)-Y(u-p).
$$
The <triangle inequality> and <operator norm> give
$$
\boxed{\|u-Yu\|\leq(1+\|Y\|)\|u-p\|
=(1+\|Y\|)E.}
$$
The estimate actually holds without attainment: apply it with any $q\in A$ instead of $p$ and take the <infimum>. Its proof applies in every <normed space>. This is <strict convexity separates uniqueness from existence in best approximation>.
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