Let $L_n:\mathcal{H}\to\mathcal{H}$ be a sequence of compact bounded linear transformation on a Hilbert space $\mathcal{H}$, and $h_m$ be a sequence in $\mathcal{H}$.
Since each $L_n$ is compact, there exists a subsequence $\{h_{nk}\}_{k=1}^\infty$ of $h_m$ such that $L_n(h_{nk})$ converges in the given Hilbert space.
Thus, we obtain a countable set $\{h_{nk}:n\in\mathbb{N},~k\in\mathbb{N}\}\subset\{h_{m}:m\in\mathbb{N}\}$.
$h_{11}~h_{12}~h_{13}~\cdots$
$h_{21}~h_{22}~h_{23}~\cdots$
$~\cdot~~~~\cdot~~~~\cdot$
$~\cdot~~~~\cdot~~~~\cdot$
$~\cdot~~~~\cdot~~~~\cdot$
Then, taking the usual Cantor's diagonalization, and then throwing away some absurd elements which prevent the resulting sequence from not being a subsequence of $h_{m}$ (as a subsequence $f_{n_k}$ of a sequence $f_n$ has to have the property that $n_k<n_{k+1}$ for all $k\in\mathbf{N}$) we can form another subsequence $h_{m_l}$ of $h_m$.
My question is that is it possible to take the subsequence $h_{m_l}$ in such a way that $L_n(h_{m_l}$ is convergent in the given space for all $n$?
First of all the existence of the subsequences $\{h_{nk}\}_{k=1}^\infty$ requires that $\{h_m\}_{m=1}^\infty$ is bounded.
The idea is to construct the subsequences $\{h_{nk}\}_{k=1}^\infty$ one after the other and not all at once. In detail:
We can find a subsequence $\{h_{1k}\}_{k=1}^\infty$ of $\{h_m\}_{m=1}^\infty$ such that $\{L_1(h_{1k})\}_{k=1}^\infty$ converges, because $L_1$ is compact. Since $\{h_{1k}\}_{k=1}^\infty$ is a subsequence of a bounded sequence it is again bounded.
And now we can find a subsequence $\{h_{2k}\}_{k=1}^\infty$ of $\{h_{1k}\}_{k=1}^\infty$ such that $\{L_2(h_{2k})\}_{k=1}^\infty$ converges and this subsequence is again bounded.
We can repeat this argument for $L_3, L_4, \ldots$ and then by diagonalization $h_{m_i} := h_{ii}$ get a subsequence such that $\{L_n(h_{m_i})\}_{i=1}^\infty$ converges for all $n$.