Minimum principle in Hilbert space

In infinite dimensional $\ell_2$, take $M=\{ \,{n+1\over n} e_n : n=1, 2,3, \ldots\,\}$, where $e_n$ is the standard $n^{\rm th}$ unit vector defined by: $e_n(m)=\cases{1,& m=n \cr 0, &\text{ otherwise}}$.

$M$ is closed, since the distance between any two of its elements is greater than $\sqrt 2$ (and thus the only convergent sequences from $M$ are those that are eventually constant). $M$ clearly is non-empty and has no element of minimal norm.

In the Hilbert space $\ell^{2}=\{x=(c_1,c_2,...)|\displaystyle\sum_{i=1}^\infty|c_i|^2<\infty\}$ with inner product $\langle x,y\rangle=\displaystyle\sum_{i=1}^\infty c_i\overline{d_i}$, we can consider $M=\ell^{2}-\{0\}$, which is not convex: since for $x\in M$, $\displaystyle\frac{x+(-x)}{2}=0\notin M$. On the other hand, there does not exist element of minimum norm, for $x_n=(1/n,0,0,...)\in M$ and $\|x_n\|=1/n$ which tends to zero as $n\rightarrow\infty$.