The trace norm cannot be increased by composing with a unitary operator $\newcommand{\tr}{\operatorname{tr}}$
I was reading a proof for the statement $|\tr(US)|\leq |\tr(S)|$, for every endomorphism $S$ on a complex vector space $H$ and every unitary operator $U$ on the same space. Though the proof is short it uses the polar decomposition and the cauchy swartz inequality. I came out with the very simple "proof":
\begin{eqnarray}
\left|\tr(US)\right| &=& \left|\tr\left(\sum_iu_{ii}|i\rangle\langle i|\sum_{jk}s_{jk}|j\rangle\langle k|\right)\right|\\
       &=& \left|\tr\left(\sum_i\sum_k u_{ii} s_{ik} |i\rangle\langle k|\right)\right| \\
       &=& \left|\sum_{i}u_{ii}s_{ii}\right|\\
       &\leq& \left|\sum_i s_{ii}\right| \\
       &=& \left|\tr(S)\right| 
\end{eqnarray}
where I just rely in the fact that unitary operators are normal and admit the spectral decomposition theorem (that is $U=\sum_i u_{ii}|i\rangle\langle i|$ for some orthonormal basis $|i\rangle$)
I would like to know if I am missing something very obvious or the proof is indeed correct.

Your statement of the "spectral decomposition theorem" is incorrect.  Unitary operators don't necessarily have point spectrum.  
Also the result is incorrect.  Try 
$$ S = U = \pmatrix{1 & 0\cr 0 & -1\cr}$$
