Suppose that $\sigma_{\mathrm{ess}}(T)=\{0\}$. So, zero is the only one possible accumulation point.
Thus if $\sigma_d(T)$ is finite then $E(\{\lambda\in\mathbb{R}:|\lambda|>\varepsilon\})$ has finite range because it is the sum of projections on finite dimensional subspaces of $H$.
In the same way, if $\sigma_d(T)$ is infinite then $E(\{\lambda\in\mathbb{R}:|\lambda|>\varepsilon\})$ has finite range because $\{\lambda:|\lambda|>\varepsilon\}$ contains only finite eigenvalues of finite multiplicity because $T$ is bounded.
