Orthonormal Sets in Hilbert Spaces let $H$ be a Hilbert space and let M be a dense linear subspace of H.
Can we find a complete orthonormal set $\{u_{\alpha}: \alpha \in A\}$ for H
in M?

I think the answer is negative in general, but I cannot find a counterexample.
Thank you very much in advance for your help.
Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis? 

It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2.

I tried to solve the exercise, but I'm not quite sure that I understood it.

Let $H_1=\ell_2(\mathbb N)$, and $H_2=\ell_2(I)$, where $I$ is a set with cardinality $\mathfrak c$ (the continuum). Choose a linearly independent family $(x_i)_{i\in I}\subset H_1$ and an orthonormal basis $(e_i)_{i\in I}$ of $H_2$. Now let $H:=H_1\oplus H_2$ and $M:={\rm span}\,\{ x_i\oplus e_i;\; i\in I\}\subset H$. Let us try to show that $M$ admits no orthonormal basis.

First, note that, identifying $H_2$ with $\{ 0\} \oplus H_2\subset H$, we have $M\cap H_2=\{ 0\}$. This is rather easy to check using that fact that the $x_i$ are linearly independent.

Now, let us show that any orthonormal family $\mathcal Z\subset M$ is countable. To see this, choose an orthonormal basis $(f_n)_{n\in\mathbb N}$ of $H_1=\ell_2(\mathbb N)\subset H$ and observe that for any $n\in\mathbb N$, we have $\langle f_n, z\rangle=0$ for all but countably many $z\in\mathcal Z$, because $\sum_{z\in\mathcal Z}\vert \langle f_n,z\rangle\vert^2&lt;\infty$. Since by the previous remark, any $z\in\mathcal Z$ must satisfy $\langle f_n ,z\rangle\neq 0$ for at least one $n\in\mathbb N$ (otherwise we would have $z\in H_1^{\perp}=H_2$ and hence $z=0$), it follows that $\mathcal Z$ must be countable. Indeed, we have $\mathcal Z=\bigcup_{n\in\mathbb N} \mathcal Z_n$, where $\mathcal Z_n=\{ z;\; \langle f_n,z\rangle\neq 0\}$ is countable for each $n$.

It follows that if $M$ were to admit an orthonormal basis, then it would be separable. But this is not possible because 
$\pi_2(M)$ is dense in $H_2$ (since it contains all the vectors $e_i$) and $H_2$ is not separable.
consider $X:=\left\{A\in\mathcal{P}\left(M\right): A \text{ is an ON-System }\right\}$ and $\subseteq$. Obviously each partially ordered chain $C$ has an upper bound in $X$: simply consider $\cup_{a\in C}a$.  By Zorn we get that there exists at least one maximal element of $X$. Take the maximal element, by definition it's an orthonormal Hilbert-basis.


  see comments. i will not delete this. (for documentation)