 Solution

Assume $K$ is nonempty; without that necessary hypothesis the assertion is false. Put $d={\inf }_{z\in K}\Vert x-z{\Vert }_p$ and choose $y_n\in K$ with $\Vert x-y_n{\Vert }_p\to d$. Convexity gives $(y_n+y_m)/2\in K$, so its distance to $x$ is at least $d$. Apply the supplied uniform-convexity inequality to $x-y_n$ and $x-y_m$:Thus the minimizing sequence is Cauchy. Completeness of the Banach space ${\ell }^p$ gives $y_n\to y$; closedness gives $y\in K$, and continuity of the norm gives $\Vert x-y{\Vert }_p=d$. The minimum distance is attained. This is the nearest point in a uniformly convex Banach space property. The same inequality applied to two minimizers shows uniqueness, though existence alone was requested.