Orthonormal eigenbasis for a compact self adjoint operator

An orthonormal family spanning $M$. For every $\lambda \in \mathrm{\Sigma }$ the eigenspace $E_{\lambda }$ is finite-dimensional by [A1], so it has an orthonormal basis by [A2]; the disjoint union $E$ of all these finite bases is an orthonormal family, because each basis is orthonormal and vectors belonging to distinct eigenvalues are orthogonal by [A1]. Its closed linear span is $M$ by the definition of $M$, and $M=(\ker T{)}^{\perp }$ by [A1].

A maximal orthonormal family in the kernel. Let $P$ be the set of orthonormal families contained in $\ker T$, ordered by inclusion. This is a nonempty poset (the empty family belongs to it) and the union of any chain in $P$ is again an orthonormal family contained in $\ker T$, hence an upper bound of the chain; therefore Zorn's lemma [A4] provides a maximal element $B\in P$.

$B$ is complete in the kernel. Let $W:=\stackrel{}{\mathrm{span}}B\subseteq \ker T$ be the closed span of $B$, which is a closed subspace of the Hilbert space $\ker T$ [A5]; if $W\mathrm{\ne }\ker T$ then by the orthogonal decomposition in the Hilbert space $\ker T$ [A3] there is $v\in \ker T\cap W^{\perp }$ with $v\mathrm{\ne }0$. Then $v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$ has norm $1$, is orthogonal to every element of $B$, and lies in $\ker T$, so $B\cup \{v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }\}$ is an orthonormal family in $\ker T$ strictly containing $B$, contradicting maximality; hence $W=\ker T$, that is $B$ is a complete orthonormal family of the Hilbert space $\ker T$.

The union is a Hilbert basis of $H$. The union $E\cup B$ is an orthonormal family: it is the union of two orthonormal families, and every $b\in B\subseteq \ker T$ is orthogonal to every $e\in E\subseteq M=(\ker T{)}^{\perp }$ by [step 1.1] and [A1]. Its closed linear span contains $M$ (by [step 1.1]) and $\ker T$ (by [step 2.1]), hence contains $M\oplus \ker T=H$ by [A1]; therefore $E\cup B$ is complete and is a Hilbert basis of $H$.

Conclusion. Every element of $E\cup B$ is an eigenvector of $T$: the vectors of $E$ lie in nonzero eigenspaces, while every $b\in B$ is a unit vector in $\ker T$, so $b\mathrm{\ne }0$ and $Tb=0=0b$ (Eigenvalues, eigenvectors, eigenspaces $E_{\lambda }(T)=\ker (T-\lambda I)$, and the spectrum ${\sigma }_F(T)$ of an endomorphism). Thus, whenever $B$ is nonempty, $0$ is an eigenvalue witnessed by each of its members; when $\ker T=\{0\}$ one has $B=\mathrm{\varnothing }$, and the proof neither needs nor asserts that $0$ is an eigenvalue. By [step 3.1] the family $E\cup B$ is a Hilbert basis of $H$ consisting of eigenvectors of $T$, which proves the corollary; the degenerate descriptions in the statement are the cases $M=\{0\}$ (then $E=\mathrm{\varnothing }$ and $B$ spans $\ker T=H$) and $\ker T=\{0\}$ (then $B=\mathrm{\varnothing }$).