I've been trying to prove that every Hilbert space has a Hilbert basis, and think I have succeeded with the proof below, but I am not sure since this would seem to suggest that it in fact has an orthonormal basis in the standard sense (that every element is a finite linear combination of basis elements), and moreover this proof only uses the inner product and not completeness. Is the proof correct and do the consequent results hold?
Let $P$ be the set of orthonormal subsets of $V,$ a pre-Hilbertian vector space and define on $P$ the partial ordering $T_1\leq T_2$ for $T_1\subset T_2.$ Then any chain $M$ in $P$ has a maximal element by the union of elements of $M,$ so Zorn's Lemma implies existence of a maximal element $B$ of $P.$
Suppose for contradiction that $B$ is not a basis, so there exists $v\in V\setminus\operatorname{span}(B).$ Choose $u\in B,$ which is non-empty for $V$ non-empty, and set $\tilde{v}=\frac{v-\langle v,u\rangle u}{\|v-\langle v,u\rangle u\|}$. Then $\|\tilde{v}\|=1$ and $\langle u,\tilde{v}\rangle =0,$ so $\{u,\tilde{v}\}$ is orthonormal, so contained in $B$ by maximality. But then $v=(\|v-\langle v,u\rangle u\|)\tilde{v}+\langle u,\tilde{v}\rangle u\in\operatorname{span}(B)$. Contradiction.
p.s. I am aware similar questions to this have been asked but I have not found my answer in any of those I have read. Standard proofs of the result I was trying to prove seem to use the orthogonal decomposition in a Hilbert space, which indeed uses the properties I avoided above.
Best Answer
There is an error in your proof, when you write that $\left\{u,\tilde v\right\}\subset B$ “by maximality”. The fact that $B$ is maximal doesn't imply that it contains every orthonormal set. In $\mathbb R^2$, endowed with its usual inner product, $\{(1,0),(0,1)\}$ is a maximal orthonormal set, $\left\{\frac35,\frac45\right\}$ is orthonormal, but $\{(1,0),(0,1)\}\not\supset\left\{\frac35,\frac45\right\}.$