Consider first a couple of special cases, where there is an obvious choice of $Z:=\operatorname{span}(z)$ with an eigenvctor $z\in Y$.

**I.** At least one of the $\beta_n$'s vanishes, say $\beta_{n_0}=0$. Then $z:=y_{n_0}$ is an eigenvector of $S$ on $Y$ .

**II.** At least two of the $\alpha_n$'s coincide, say $\alpha_{n_0}=\alpha_ {n_1}=\lambda$, for $n_0\neq n_1$. Then some non-trivial linear combination $p\beta_{n_0}+q\beta_{n_1}$ vanishes; $z:= p y_{n_0}+qy_{n_1}$ is an eigenvector with eigenvalue $\lambda$.

Conversely, if there is such an invariant subspace $Z\subset Y$, then the compact operator $S_{|Z}$ has an eigenvalue $\lambda$ with eigenvector $z:=\sum_{n=1}^\infty c_ny_n\in Y$. This means $$\sum_{n=1}^\infty \alpha_n c_ny_n + \left(\sum_{n=1}^\infty \beta_nc_n\right)x_0=\lambda \sum_{n=1}^\infty c_ny_n\\ ,$$
so $(\alpha_n-\lambda)c_n=0$ for all $n$ and $\sum_{n=1}^\infty \beta_nc_n=0$. Hence, if only one $c_n$ is non-zero, we are in case **I**; if there are at least two non-zero $c_n$, we are in case **II.**