**Step 1: $\sigma(T) \subset \overline{B}(0, \|T\|)$.** If $|\lambda| > \|T\|$, then $\|\lambda^{-1}T\| < 1$, so $I - \lambda^{-1}T$ is invertible by the Neumann series $\sum_{k=0}^\infty (\lambda^{-1}T)^k$. Therefore $(T - \lambda I) = -\lambda(I - \lambda^{-1}T)$ is invertible, giving $\lambda \in \rho(T)$. **Step 2: $\rho(T)$ is open (equivalently, $\sigma(T)$ is closed).** Fix $\lambda_0 \in \rho(T)$. For $\mu \in \mathbb{C}$, write $T - \mu I = (T - \lambda_0 I) - (\mu - \lambda_0)I = (T - \lambda_0 I)[I - (\mu - \lambda_0)(T - \lambda_0 I)^{-1}]$. If $|\mu - \lambda_0| < \|(T - \lambda_0 I)^{-1}\|^{-1}$, the Neumann series shows $I - (\mu - \lambda_0)(T - \lambda_0 I)^{-1}$ is invertible, so $T - \mu I$ is invertible and $\mu \in \rho(T)$. **Step 3: $\sigma(T) \ne \varnothing$.** Suppose $\sigma(T) = \varnothing$, so the resolvent $R(\lambda) := (T - \lambda I)^{-1}$ is defined for all $\lambda \in \mathbb{C}$. The Neumann series shows $R$ is analytic (given by a convergent [power series](/page/Power%20Series) in a neighbourhood of every $\lambda_0 \in \mathbb{C}$). For $|\lambda| > \|T\|$: \begin{align*} \|R(\lambda)\| = \|\lambda^{-1}(I - \lambda^{-1}T)^{-1}\| \le \frac{1}{|\lambda| - \|T\|} \to 0 \text{ as } |\lambda| \to \infty. \end{align*} So $R$ is a bounded entire $\mathcal{L}(X)$-valued function. By Liouville's theorem for Banach-space-valued [functions](/page/Function) (which uses the Hahn-Banach theorem: for every $f \in \mathcal{L}(X)^*$, $f \circ R$ is a bounded entire scalar function, hence constant by the classical Liouville theorem, hence $R$ is constant). But $R(\lambda) \to 0$ as $|\lambda| \to \infty$, so $R \equiv 0$, contradicting $R(\lambda)(T - \lambda I) = I$.