Note for positive sums:
$$a_{\alpha\beta}\geq0:\quad\sum_\alpha\sum_\beta a_{\alpha\beta}=\sum_\beta\sum_\alpha a_{\alpha\beta}$$
By Parseval one has:
$$\sum_\sigma\langle|A|\sigma,\sigma\rangle=\sum_\sigma\||A|^{1/2}\sigma\|^2=\sum_\sigma\sum_\tau|\langle|A|^{1/2}\sigma,\tau\rangle|^2\\
=\sum_\tau\sum_\sigma|\langle\sigma,|A|^{1/2}\tau\rangle|^2=\sum_\tau\||A|^{1/2}\tau\|^2=\sum_\tau\langle|A|\tau,\tau\rangle$$
Concluding independence.
I like your idea of building a sequence from the orthonormal basis vectors and proving non-existence of convergent subsequences! But I think we need to be a bit more restrictive in how we select our sequence: rather than selecting all the orthonormal basis vectors to be our sequence, we should select only the "troublesome" ones.
So here is the idea. Suppose, for contradiction, that $\beta_n$ does NOT tend to zero. Then hopefully you can show that there exists a $\epsilon > 0$ and there exists an ascending sequence $n_1, n_2, n_3, \dots$ such that
$$ |\beta_{n_1}| > \epsilon, \ \ \ | \beta_{n_2} | > \epsilon, \ \ \ | \beta_{n_3} | > \epsilon \dots $$
Now consider the sequence
$$ e_{n_1}, \ \ e_{n_2}, \ \ e_{n_3}, \ \ \dots$$
Hopefully you can verify that the sequence $$F(e_{n_1}), F(e_{n_2}), F(e_{n_3}), \dots$$ cannot possibly contain a Cauchy subsequence, which would be enough to complete your proof.
To see the problem with using the entire orthonormal basis $e_1, e_2, e_3, \dots$ as the "test" sequence in your argument: Consider the example where $\beta_n$ is $1$ when $n$ is odd and $0$ when $n$ is even. Then $F(e_n)$ clearly contains a convergent subsequence, namely, $F(e_2), F(e_2), F(e_6), \dots$ So we don't get a contradiction. We need to be more restrictive, by picking only the "troublesome" elements $e_1, e_3, e_5, \dots$ to be our "test" sequence, and only then do we find ourselves unable to find a Cauchy subsequence.
Best Answer
Here's a different proof.
Assume first that $A$ is finite-rank. Then $\text{Tr}(A^*A)<\infty$, and so $$ 0\leq\text{ Tr}(A^*A)=\sum_{n=1}^\infty\langle A^*Ae_n,e_n\rangle=\sum_{n=1}^\infty\|Ae_n\|^2<\infty, $$ and so $\|Ae_n\|\to0$.
If $A$ is any compact operator, there exists a sequence of finite-rank operators $\{A_m\}$ with $\|A_m- A\|\to0.$ Then $$ \|Ae_n\|\leq\|(A-A_m)e_n\|+\|A_me_n\|\leq\|A_m-A\|+\|A_me_n\|. $$ So $$ 0\leq\limsup_n\|Ae_n\|\leq\|A_m-A\|+0=\|A_m-A\|. $$ As $m$ was arbitrary, we conclude that $0\leq\limsup_n\|Ae_n\|=0$, and so $\lim_n\|Ae_n\|=0$.