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 $\lim_n\|Ae_n\|=0$.
No, we cannot conclude that the operator is trace class.
For example, let a Hilbert space have orthonormal basis $e_1,f_2,e_2,f_2,e_3,f_3,\ldots$, and $T$ interchanges $e_i,f_i$, while multiplying both by a positive real $\lambda_i$. That is, in these coordinates, the matrix of $T$ is a list of diagonal blocks, with the $i$-th diagonal block being anti-diagonal $\lambda_i,\lambda_i$.
For $\lambda_i\rightarrow 0$, the operator is compact, almost from the definition.
All the diagonal entries are $0$.
The operator is self-adjoint because the matrix is symmetric real.
However, the operator is not trace class unless $\sum_i |\lambda_i|<\infty$, which easily fails for many sequences of positive reals $\lambda_i\rightarrow 0$.
Edit: It is noteworthy that the analogous characterization (I pointedly don't say "definition") of "Hilbert-Schmidt" does not depend on choice of basis. Thus, "defining" trace-class as composition of two Hilbert-Schmidt operators is sometimes usefully more intrinsic, less basis/coordinate-dependent.
Best Answer
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.