Is convexity necessary for the existence of a shortest-norm vector?

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There's a well-known result in the theory of Hilbert spaces, namely the projection on closed convex sets:

If $C$ is a nonempty closed convex set in a Hilbert space $H$, then there exists a unique vector in $C$ that has the smallest norm.

Now, I understand the oft-stated proof which uses the parallelogram law. However, I was expecting that convexity is required only for the proof of uniqueness (one easily thinks of annuli around the origin) and that existence was a result only of closed-ness. But even the existence argument in the proof uses convexity.

Thus I ask of a counterexample to my intuition, namely a Hilbert space in which a closed nonempty set has no vector of least norm.


Okay, so I can understand that if the space is infinite-dimensional, then we can take the set containing $(1 + 1/n)e_n$ where $e_n$'s are orthonormal.

This raises the question: What about finite-dimensional spaces?