[proofplan]
We represent the normal operator $T$ by its spectral measure $E$ and use the support characterization of the spectrum. Since $\lambda\in\sigma(T)=\operatorname{supp}E$, every small spectral neighbourhood of $\lambda$ has nonzero spectral projection. Choosing unit vectors in these spectral subspaces localizes their scalar spectral measures near $\lambda$, and the [Borel functional calculus](/theorems/2696) then estimates $\|(T-\lambda I)h_n\|_H$ by the radius of the neighbourhood.
[/proofplan]
[step:Represent $T$ by its spectral measure and identify $\lambda$ as a support point]
By the [citetheorem:8412], there is a projection-valued measure
\begin{align*}
E:\mathcal B(\sigma(T))\to\mathcal L(H)
\end{align*}
such that
\begin{align*}
T=\int_{\sigma(T)} z\,dE(z).
\end{align*}
By [citetheorem:8414],
\begin{align*}
\sigma(T)=\operatorname{supp}E.
\end{align*}
Fix $\lambda\in\sigma(T)$. For each $n\in\mathbb N$, define the relatively open Borel set
\begin{align*}
D_n:=\sigma(T)\cap B(\lambda,1/n).
\end{align*}
Since $\lambda\in\operatorname{supp}E$ and $D_n$ is a neighbourhood of $\lambda$ in $\sigma(T)$, the projection $E(D_n)$ is nonzero.
[guided]
The first task is to convert the spectral assertion $\lambda\in\sigma(T)$ into a supply of vectors. Since $T$ is normal on a complex [Hilbert space](/page/Hilbert%20Space), the [citetheorem:8412] gives a projection-valued measure
\begin{align*}
E:\mathcal B(\sigma(T))\to\mathcal L(H)
\end{align*}
with
\begin{align*}
T=\int_{\sigma(T)} z\,dE(z).
\end{align*}
The support theorem [citetheorem:8414] identifies the spectrum with the support of this spectral measure:
\begin{align*}
\sigma(T)=\operatorname{supp}E.
\end{align*}
Now fix $\lambda\in\sigma(T)$. For $n\in\mathbb N$, define
\begin{align*}
D_n:=\sigma(T)\cap B(\lambda,1/n).
\end{align*}
This is a Borel subset of $\sigma(T)$ and is a relatively open neighbourhood of $\lambda$ in $\sigma(T)$. The meaning of $\lambda\in\operatorname{supp}E$ is precisely that every such neighbourhood has nonzero spectral projection. Hence
\begin{align*}
E(D_n)\ne 0
\end{align*}
for every $n\in\mathbb N$. This nonzero projection is the source of the approximate eigenvectors: vectors in its range are spectrally concentrated within distance $1/n$ of $\lambda$.
[/guided]
[/step]
[step:Choose unit vectors supported in shrinking spectral neighbourhoods]
For each $n\in\mathbb N$, since $E(D_n)\ne 0$, its range is a nonzero closed subspace of $H$. Choose $g_n\in\operatorname{Range}(E(D_n))$ with $g_n\ne 0$, and define
\begin{align*}
h_n:=\frac{g_n}{\|g_n\|_H}.
\end{align*}
Then $\|h_n\|_H=1$ and, because $E(D_n)$ is a projection and $g_n\in\operatorname{Range}(E(D_n))$,
\begin{align*}
E(D_n)h_n=h_n.
\end{align*}
[/step]
[step:Localize the scalar spectral measure of each chosen vector]
For each $n\in\mathbb N$, define the scalar spectral measure
\begin{align*}
\mu_n:\mathcal B(\sigma(T))\to\mathbb C
\end{align*}
by
\begin{align*}
\mu_n(\Delta):=(E(\Delta)h_n,h_n)_H.
\end{align*}
Since $E(D_n)h_n=h_n$ and projection-valued measures satisfy $E(\Delta)E(D_n)=E(\Delta\cap D_n)$, we have
\begin{align*}
\mu_n(\sigma(T)\setminus D_n)=(E(\sigma(T)\setminus D_n)E(D_n)h_n,h_n)_H=0.
\end{align*}
Thus $\mu_n$ is supported in $D_n$. Also,
\begin{align*}
\mu_n(\sigma(T))=(E(\sigma(T))h_n,h_n)_H=\|h_n\|_H^2=1.
\end{align*}
[/step]
[step:Estimate the resolvent defect by the spectral localization]
Define the bounded Borel function
\begin{align*}
f:\sigma(T)\to\mathbb C,\qquad z\mapsto z-\lambda.
\end{align*}
By the Borel functional calculus rules in [citetheorem:8413],
\begin{align*}
T-\lambda I=f(T)
\end{align*}
and
\begin{align*}
(T-\lambda I)^*(T-\lambda I)=|f|^2(T).
\end{align*}
Therefore, using the scalar spectral integral formula,
\begin{align*}
\|(T-\lambda I)h_n\|_H^2=\int_{\sigma(T)} |z-\lambda|^2\,d\mu_n(z).
\end{align*}
Since $\mu_n$ is supported in $D_n$ and $|z-\lambda|<1/n$ for every $z\in D_n$,
\begin{align*}
\|(T-\lambda I)h_n\|_H^2\le \frac{1}{n^2}\mu_n(D_n)\le \frac{1}{n^2}.
\end{align*}
Hence
\begin{align*}
\|(T-\lambda I)h_n\|_H\le \frac{1}{n}.
\end{align*}
[/step]
[step:Conclude that $\lambda$ is an approximate eigenvalue]
The sequence $(h_n)_{n=1}^{\infty}$ satisfies $\|h_n\|_H=1$ for every $n\in\mathbb N$ and
\begin{align*}
\lim_{n\to\infty}\|(T-\lambda I)h_n\|_H=0
\end{align*}
by the estimate $\|(T-\lambda I)h_n\|_H\le 1/n$. Therefore $\lambda$ is an approximate eigenvalue of $T$. Since $\lambda\in\sigma(T)$ was arbitrary, every spectral value of $T$ is an approximate eigenvalue.
[/step]
