I wonder if there was an easier way than proving Lidskii's theorem in order to show, for the particular case of $U$, compact trace class operator which is also self-adjoint to show that its eigenvalues $\displaystyle \left(\lambda_n\right)_{n \in \mathbb{N}}$ $$ \sum_{n=0}^{+\infty}\lambda_n=\text{Tr}\left(U\right) $$
I was trying to prove it using linear span of the hilbert's basis and let $n \rightarrow +\infty$, isn't this possible ?
Since $U$ is selfadjoint, it is a standard computation to show that distinct eigenvalues have orthogonal eigenspaces. The fact that $U$ is trace-class guarantees that all eigenspaces (with the possible exception of zero) are finite-dimensional.
When the above is put together, you get the Spectral Theorem: there exists an orthonormal basis $\{e_n\}$ such that $$ U=\sum_n\lambda_n\langle \cdot,e_n\rangle\,e_n. $$ And now you can calculate the trace using that particular orthonormal basis: since $Ue_n=\lambda_ne_n$, $$ \operatorname{Tr}(U)=\sum_n\langle Ue_n,e_n\rangle=\sum_n\langle\lambda_ne_n,e_n\rangle=\sum_n\lambda_n. $$