Here, the concept of "orthonormal Hamel basis" of an innper product space $V$ is a family of orthonormal family $\{e_i\}_{i\in I}$ (may be uncountable) such that any $x\in V$ is a finite combination of elements from $\{e_i\}_{i\in I}$.
It is well-known that any Hilbert space couldn't have such orthonormal Hamel basis. Indeed, let $\{e_i\}_{i=1}^n$ be a sequence of $\{e_i\}_{i\in I}$ and consider the element $$ f=\sum_{i=1}^\infty \frac{1}{i} e_i. $$ Obviously, $f$ can't be spanned by any finite combination of $\{e_i\}_{i\in I}$.
On the other hand, some inner product space may be spanned by an "orthonormal Hamel basis. Say, the polynomial space $\mathbb{P}[0,1]$ has an countable orthonormal basis. It is well-kown that it is complted and any infinite dimensional subspace of it also fails to be compelted. Does it mean that it can also be (finitely) spanned by some orthonormal sets. So, I even wonder that
if an inner product space has an "orthonormal Hamel basis", then does any subspace of it has an "orthonormal Hamel bais?"
My guess is No. Any counterexample? If $\{e_i\}_{i\in I}$ is countalbe, then this is true. Because, we can introduce $V_n=span\{e_1,\ldots,e_n\}$. Let $X$ be any subspace of $V$. Then we have $$ X=\cup_{n=1}^\infty (V_n\cap X). $$ We can easily find the bases by Gram-schmidt process.