Spectral decomposition of compact self-adjoint operator

What he means is that you start with $e_1$ (assume that is nonzero, otherwise you start with a bigger index). Then you have an orthonormal basis $$\tag1 \xi_{1,1},\ldots,\xi_{1,m_1} $$ of $e_1\mathfrak H$. As $e_2\geq e_1$, you take an orthonormal basis of $e_2\mathfrak H$ formed by expading $(1)$ to an orthonormal basis. That is, $$\tag2 \xi_{2,1},\ldots,\xi_{2,m_2}=\xi_{1,1},\ldots,\xi_{1,m_1},\xi_{2,m_1+1},\ldots,\xi_{2,m_2}. $$ So in each step you are enlarging the orthonormal family, and when you consider all $n$ you denote it by $\{\xi_n\}$. For each $n$ you have $$ xe_n=\sum_{k=1}^{m_n}\alpha_{n,k}\,t_{\xi_{n,k},\xi_{n,k}}. $$ The way the elements were constructed in $(2)$ guarantees that $\xi_{n+r,k}=\xi_{n,k}$ and $\alpha_{n+r,k}=\alpha_{n,k}$ if $k\leq m_n$. So, for $n$ big enough, the $\xi_{n,k}$ and the $\alpha_{n,k}$ do not depend on $n$. That, together with $x=\lim_n xe_n$, allow us to write $$ x=\sum_k\alpha_k\,t_{\xi_k,\xi_k}. $$