Selfadjoint compact operator with finite trace

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.

Disclaimer: Non-Compact Operators!

Given the Hilbert space $\ell^2(\mathbb{N})$.

Consider sum of shifts:* $$A_\pm:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A_\pm:=R\pm L$$

They have finite trace: $$\sum_n\langle A_\pm\delta_n,\delta_n\rangle=\sum_n0=0$$

But for the shifts: $$\sum_n\langle|R|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ $$\sum_n\langle|L|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ Thus for the sum: $$\operatorname{Tr}A_\pm<\infty\implies\operatorname{Tr}A_\mp<\infty$$ Concluding counterexample.

*Shifts: Right & Left