every Hilbert space has an orthonormal basis

Theorem - Every Hilbert space $H\ne \{0\}$ has an orthonormal basis.

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Proof : As could be expected, the proof makes use of Zorn’s Lemma. Let $𝒪$ be the set of all orthonormal sets of $H$. It is clear that $𝒪$ is non-empty since the set $\{x\}$ is in $𝒪$, where $x$ is an element of $H$ such that $\parallel x\parallel =1$.

The elements of $𝒪$ can be ordered by inclusion, and each chain $𝒞$ in $𝒪$ has an upper bound, given by the union of all elements of $𝒞$. Thus, Zorn’s Lemma assures the existence of a maximal element $B$ in $𝒪$. We claim that $B$ is an orthonormal basis of $H$.

It is clear that $B$ is an orthonormal set, as it belongs to $𝒪$. It remains to see that the linear span of $B$ is dense in $H$.

Let $\overline{\mathrm{span}B}$ denote the closure of the span of $B$. Suppose $\overline{\mathrm{span}B}\ne H$. By the orthogonal decomposition theorem we know that

$$H=\overline{\mathrm{span}B}\oplus {(\overline{\mathrm{span}B})}^{⟂}$$

Thus, we conclude that ${(\overline{\mathrm{span}B})}^{⟂}\ne \{0\}$, i.e. there are elements which are orthogonal (http://planetmath.org/OrthogonalVectors) to $\overline{\mathrm{span}B}$. This contradicts the maximality of $B$ since, by picking an element $y\in {(\overline{\mathrm{span}B})}^{⟂}$ with $\parallel y\parallel =1$, $B\cup \{y\}$ would belong belong to $𝒪$ and would be greater than $B$.

Hence, $\overline{\mathrm{span}B}=H$, and this finishes the proof. $\mathrm{\square }$