Existence of a maximal orthonormal family, and maximality as completeness

The family of orthonormal subsets of $H$, ordered by inclusion, is a nonempty poset: the empty set is orthonormal. Every chain of orthonormal sets has an upper bound, namely its union, which is orthonormal by [A2] and contains each member of the chain.

Complete orthonormal sets are maximal. Suppose the closed linear span of an orthonormal set $S$ is $H$ and let $T\supseteq S$ be orthonormal. If $t\in T\setminus S$, then $t$ is orthogonal to every element of $S$ because $T$ is orthonormal; the vector $t$ also lies in the closed linear span of $S$, so $t$ is orthogonal to itself by [A5], whence $\mathrm{\parallel }t{\mathrm{\parallel }}^2=⟨t,t⟩=0$ and $t=0$, contradicting $\mathrm{\parallel }t\mathrm{\parallel }=1$. Thus no orthonormal set strictly contains $S$, so $S$ is maximal.

Non-complete orthonormal sets are not maximal. Suppose the closed linear span of an orthonormal set $S$, indexed by itself, is not $H$. By the equivalence of completeness with the vanishing of the orthogonal complement, $S^{\perp }\mathrm{\ne }\{0\}$, so there is $z\mathrm{\ne }0$ orthogonal to every element of $S$; then $u:=z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ has $\mathrm{\parallel }u\mathrm{\parallel }=1$ and is orthogonal to every element of $S$, so $S\cup \{u\}$ is orthonormal and strictly larger than $S$. Hence $S$ is not maximal.

By Zorn's lemma applied to the poset of orthonormal subsets, whose chains are bounded by step 1.1, there is an orthonormal set $S_0\subseteq H$ maximal under inclusion, which is claim 1.

Steps 1.2 and 1.3 prove that an orthonormal set is maximal exactly when it is complete, which is claim 2; the maximal set $S_0$ is then complete. Indexing a complete orthonormal set by itself gives a complete orthonormal family and hence a Hilbert basis, and an orthonormal family whose image is maximal is complete by claim 2, which is claim 3.