Characterising Hilbert spaces I found this question:
Characterization of Hilbert spaces using orthogonal decomposition
It deals with the question of when pre-Hilbert spaces are Hilbert. It proves that (assuming separability)

A pre-Hilbert space $H$ is Hilbert if for all subspaces $K$
$$K^\perp = \{0\} \implies K\; \text{dense in}\; H $$

But I'm afraid I do not really understand the answer. In particular I would have thought that neither $\eta$ nor $\xi$ lie in $K$.
Additionally I would very much like a proof in the non-separable case. There is a comment saying it should be possible, but again I cannot fill in the details.
Thanks in advance!
 A: Let $H$ be a pre-Hilbert space, denote with $\mathcal H$ its completion which is Hilbert. For a subset $K$ of $H$ denote with $K'$ the orthogonal complement in $H$ and $K^\perp$ the orthogonal complement in $\mathcal H$. Its clear that $K'\subseteq K^\perp$ and even that $K'=K^\perp \cap H$.
Suppose there is a vector $v\in \mathcal H - H$ (ie that $H$ is not Hilbert), then the span $V:=\Bbb C\cdot v$ is a closed sub-space of $\mathcal H$ and define:
$$K := V^\perp \cap H.$$
Note that since $H$ is dense in $\mathcal H$ you have that $K$ is dense in $V^\perp$ in $\mathcal H$. In particular you have that
$$K^\perp = (\overline K)^\perp = (V^\perp)^\perp = V$$
since $V$ is closed. But $K'= K^\perp\cap H = V\cap H = \{0\}$. However $K$ cannot be dense in all of $H$, because then it must be dense in $\mathcal H$ and then $V^\perp$ must be dense in $\mathcal H$. But $v$ itself cannot be approximated by elements of $V^\perp$.
This shows
$$[\text{$H$ is not Hilbert}]\implies [\text{there exists a not dense subspace $K$ with $K\neq H$ and $K^\perp = \{0\}$}]$$
independently of any countability assumptions.
The other direction is elementary as $(K^\perp)^\perp = \overline K$ in a Hilbert space, and $\{0\}^\perp$ is always the entire space.
