= Solution
The statement requires $K\ne\varnothing$; an empty set is closed and convex but contains no minimizer. Assume this necessary condition and put $d=\inf_{g\in K}\|g-f\|_p$. Since $K$ is closed and $f\notin K$, $d>0$. Choose a <minimizing sequence> $g_j\in K$ with $\|g_j-f\|_p\to d$.
The midpoint $(g_j+g_k)/2$ lies in the <convex set> $K$. Applying the stated <Clarkson inequality> to $g_j-f$ and $g_k-f$ gives
$$
\|g_j-g_k\|_p^p
\leq2^{p-1}\bigl(\|g_j-f\|_p^p+\|g_k-f\|_p^p\bigr)
-\|g_j+g_k-2f\|_p^p
\leq2^{p-1}\bigl(\|g_j-f\|_p^p+\|g_k-f\|_p^p\bigr)-2^pd^p.
$$
The right side tends to zero, so $g_j$ is a <Cauchy sequence>. By <completeness of Lp spaces>, proved in Question 2(a), it converges to some $h$. Closedness puts $h$ in $K$, and continuity of the norm gives $\boxed{\|h-f\|_p=d}$. This is the <closest point theorem for a closed convex subset of Lp>. The same inequality applied to two minimizers forces their difference to be zero, so the minimizer is also unique.
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