It is known that if $E$ is a dense subspace of a separable Hilbert space $H$, then there exists an orthonormal basis of $H$ which is contained in $E$.
I don't believe this is true for just a dense subset $E$, not a subspace. What if $E$ has the additional property that for any non-zero $\alpha\in\mathbb{C}$, $\alpha E=E$. In other words $E$ is closed under scalar multiplication, but not under addition?
Here is a counter example in $\Bbb{C}^2$.
Fix an irrational number $u \in \Bbb{R} - \Bbb{Q}.$ For $r, s \in \Bbb{R}_{>0}$ and $n,m \in \Bbb{Z}$ define $$v_{r,s,n,m} = (re^{\pi inu}, se^{\pi imu}) \in \Bbb{C}^2.$$ Let's call these "spanning vectors". Clearly the spanning vectors are dense in $\Bbb{C}^2$. Let $E$ be the union of the 1-dimensional vectorspaces spanned by the spanning vectors: $$E = \{\alpha v_{r,s,n,m}| r,s \in \Bbb{R}_{>0}, n,m \in \Bbb{Z}, \alpha \in \Bbb{C}\}.$$ Since $E$ is a union of vectorspaces $E$ is closed under multiplication. And since $E$ contains all spanning vectors we also have that $E$ is dense in $\Bbb{C}^2$. However $E$ can contain no orthonormal basis since no two spanning vectors are orthogonal. Indeed $$(v_{r_0,s_0,n_0,m_0}, v_{r_1,s_1,n_1,m_1} ) = r_0r_1e^{\pi i u(n_0-n_1) } + s_0s_1e^{\pi i u(m_0-m_1)}.$$ If this was $0$ we would have $$e^{\pi i u(m_0-m_1 - n_0 + n_1)} = - \frac{r_0r_1}{s_0s_1}.$$ But this is only possible if $ - \frac{r_0r_1}{s_0s_1} = -1$ and $u(m_0-m_1 - n_0 + n_1)$ is an odd integer, which is clearly not the case since we chose $u$ to be irrational.