= Solution
Work with bounded <self-adjoint operators> on separable infinite-dimensional <Hilbert spaces>, the setting in which the classification holds. If $UTU^*-S$ is compact for a unitary $U$, their images in the <Calkin algebra> have identical spectra, by compact perturbation invariance and unitary conjugation. Thus their essential spectra coincide.
Conversely apply the <Weyl-von Neumann theorem> to write $T=D_a+K_a$ and $S=D_b+K_b$ in <orthonormal bases>, with bounded real diagonal sequences and compact errors. The essential spectrum of a diagonal operator is exactly its sequence cluster set. Indeed outside that cluster set all but finitely many entries stay a positive distance from $\lambda$, so the diagonal has an inverse on a finite-codimensional subspace and is <Fredholm>. At a cluster point, infinitely many distinct basis vectors have $(D-\lambda)e_j\to0$; these are weakly null unit vectors, contradicting any <Fredholm> parametrix. This also handles a value with infinite multiplicity.
The two diagonal sequences therefore have the same cluster set. Part (c) provides permutations with $a_{m_k}-b_{n_k}\to0$. Send the $m_k$th basis vector for $T$ to the $n_k$th basis vector for $S$. Under this unitary, the diagonal difference has entries tending to zero and is compact, by finite-rank truncation. Adding the two conjugated compact errors gives \b[$\boxed{UTU^*-S\text{ compact}}$].
Without separability, equality of essential spectra need not suffice, even on the same <Hilbert space>. Let $H$ have uncountable dimension and let $T,S$ be projections with infinite-dimensional kernels, but with range dimensions respectively countable and uncountable. Both essential spectra are $\{0,1\}$. After any unitary conjugation the range of $T$ is still separable. If its difference from $S$ were compact, the range of $S$ would lie in the sum of that separable range and the separable closure of the compact error's range, which is impossible. Finite-dimensional spaces also need a separate dimension convention, since essential spectra are then empty. These are genuine qualifications to a statement made without the standard separable infinite-dimensional setting.
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