Let $H$ a Hilbert space and $(e_n) \subset H$ an orthonormal sequence.
Define $T: H \rightarrow H$ such that $T(x) = \sum_{k=0}^{\infty}c_k \langle x,e_k \rangle e_k$ where $c_k \in \mathbb{C}$.
I need to prove that if $T$ is compact then $c_n \rightarrow 0$.
I started by definition, $(e_n)$ is orthonormal, hence bounded so plugging this sequence in $T$ seems the natural thing to do. So we get $T(e_n)=\sum_{k=0}^{\infty}c_k \langle e_n,e_k \rangle e_k = c_n \langle e_n, e_n \rangle e_n = c_ne_n$ and by the compactness assumption $T(e_n)$ has a cauchy subsequence $c_{n_k} e_{n_k}$, which is also convergent since $H$ is complete. I can prove that $c_{n_k}$ is convergent using the limit definition, though not sure how much this helps.
How can we prove $c_n \rightarrow 0$?
Notice that, for all $m\ne n$, $\lVert c_{n}e_{n}-c_{m}e_{m}\rVert=\sqrt{\lvert c_{n}\rvert^2+\lvert c_{m}\rvert^2}\ge\lvert c_n\rvert$, therefore any Cauchy subsequence of $\{c_ne_n\}_{n\in\Bbb N}$ satisfies $c_{n_k}\to 0$.
With this in mind, let's prove that $c_n\to 0$ by showing that every subsequence $\{c_{n_k}\}_{k\in\Bbb N}$ has a sub-subsequence $\left\{c_{n_{k_h}}\right\}_{h\in\Bbb N}$ such that $c_{n_{k_h}}\to 0$. Let $\left\{c_{n_k}\right\}_{k\in\Bbb N}$ be a subsequence. The sequence $\{c_{n_k}e_{n_k}\}_{k\in\Bbb N}$ has a Cauchy subsequence $\left\{c_{n_{k_h}}e_{n_{k_h}}\right\}_{h\in\Bbb N}$ by compactness of $T$. Therefore $c_{n_{k_h}}\to 0$ by the previous observation.