I would like to show that all non-zero eigenvalues of a (possibly unbounded) compact and hermitian operator $T:\mathcal{D}(T)\to \mathcal{H}$ on a seperable, infinite-dimensional Hilbert space have finite multiplicity.
My idea so far: I would like to use the characterisation:
$\lambda\in\sigma (T)$ is eigenvalue with finite multiplicity if and only if $\lambda\in\sigma (T)$ and for all sequences $(x_{n})$ in $\mathcal{D}(T)$ with $\Vert x_{n}\Vert=1$ and $\Vert (T-\lambda )x_{n}\Vert\to 0$ there exists a convergent subsequence.
So let $\lambda\in\sigma (T)$ be non-zero and $(x_{n})$ be such a sequence. Since this sequence is by assumption bounded, there exists a weakly-convergent subsequence by Banach-Alaoglu, so lets write $x_{n_{k}}\rightharpoonup x_{0}\in\mathcal{H}$. By compactness of $T$, it follows that $Tx_{n_{k}}\to Tx_{0}$. But now I don't know how to proceed. Maybe I can show that $\Vert x_{n_{k}}\Vert\to \Vert x_{0}\Vert$, because than it would follow that $x_{n_{k}}\to x_{0}$....I haven't use the assumptions $\Vert (T-\lambda )x_{n}\Vert\to 0$ and $\lambda\neq 0$ yet....
If $N=\mathcal{N}(T-\lambda I)$ is infinite-dimensional for some $\lambda\ne 0$, then there exists a sequence $\{ x_n \}$ of unit vectors that are mutually orthogonal and satisfy $(T-\lambda I)x_n=0$. Then $T$ maps the bounded sequence $\{ x_n \}$ to a sequence $\{ \lambda x_n \}$ with no convergent subsequence. So $\mathcal{N}(T-\lambda I)$ must be finite-dimensional for all $\lambda\ne 0$.