= Solution
For the <Fredholm> convention, define
$$
\boxed{\sigma_{\mathrm{ess}}(T)=\{\lambda\in\mathbb C:T-\lambda I\text{ is not Fredholm}\}
=\sigma_{\mathcal Q(H)}([T]).}
$$
The equality follows from the <Atkinson theorem>. The quotient is a unital complex <Banach algebra> when $H$ is infinite-dimensional. Its invertible set is open by the <Neumann series>, so the spectrum is closed, and it lies in the disk $|\lambda|\leq\|[T]\|$.
For nonemptiness, suppose the quotient resolvent $R(\lambda)=(\lambda1-[T])^{-1}$ existed for every complex $\lambda$. It is analytic everywhere, and outside a sufficiently large disk its Neumann expansion gives $\|R(\lambda)\|\leq1/(|\lambda|-\|[T]\|)$. Continuity bounds it on the remaining compact disk. Every continuous linear functional applied to $R$ is therefore a bounded entire scalar function, constant by Liouville's theorem and zero because it tends to zero at infinity. The Hahn-Banach theorem, or separation by the dual, forces $R(\lambda)=0$, contradicting its inverse identity. \b[The essential spectrum is nonempty and closed.]
Infinite dimension is necessary: every operator on a finite-dimensional <Hilbert space> is <Fredholm>, so its essential spectrum in this convention is empty. The proof works on nonseparable infinite-dimensional spaces as well; it needs neither self-adjointness nor a sequence basis.
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