Proof.

1. Note that $$\[ {\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}-{ \mathopen{}\left({λ}_{0}-T\right)\mathclose{} }^{-1}= {\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}\mathopen{}\left(\mathopen{}\left({λ}_{0}-T\right)\mathclose{}-\mathopen{}\left(λ-T\right)\mathclose{}\right)\mathclose{}{ \mathopen{}\left({λ}_{0}-T\right)\mathclose{} }^{-1}= \mathopen{}\left({λ}_{0}-λ\right)\mathclose{}{\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}{ \mathopen{}\left({λ}_{0}-T\right)\mathclose{} }^{-1}\text{,} \]$$ divide by $\( λ-{λ}_{0} \)$ and use continuity of inversion.
2. For $\( \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\gt \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \)$, $\( {\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}= { \mathopen{}\left(λ\mathopen{}\left(1-\frac{1}{λ}T\right)\mathclose{}\right)\mathclose{} }^{-1}= \frac{1}{λ}{ \mathopen{}\left(1-\frac{1}{{λ}_{1}}T\right)\mathclose{} }^{-1} \)$. Then $$\[ \mathopen{}\left\lVert{}{\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}\right\rVert\mathclose{}= \frac{1}{\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}}\mathopen{}\left\lVert{}{ \mathopen{}\left(1-\frac{1}{λ}T\right)\mathclose{} }^{-1}\right\rVert\mathclose{}\leq \frac{1}{\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}}\frac{1}{1-\frac{\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}{\mathopen{}\left\lVert{}λ\right\rVert\mathclose{}}}= \frac{1}{\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}-\mathopen{}\left\lVert{}T\right\rVert\mathclose{}} \text{.} \]$$

Proof. Suppose $\( T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{} \)$ has empty spectrum. That is, $\( λ-T\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1} \)$ for all $\( λ\in \mathbb{C} \)$. Take $\( φ\in {\mathcal{L}\mathopen{}\left( X\right)\mathclose{}}^{*} \)$ and consider $\( f\mathopen{}\left( λ\right)\mathclose{}= φ\mathopen{}\left( { \mathopen{}\left(λ-T\right)\mathclose{} }^{-1}\right)\mathclose{} \)$. It follows from Proposition II.40 and the continuity of $\( φ \)$ that $\( f \)$ is analytic on $\( \mathbb{C} \)$ (with derivative $\( f' \mathopen{}\left( λ\right)\mathclose{}= {-} φ\mathopen{}\left( {\mathopen{}\left(λ-T\right)\mathclose{}}^{{-}2}\right)\mathclose{} \)$). Furthermore, $\( \lim_{\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\to\infty}{}\mathopen{}\left\lvert{}f\mathopen{}\left( λ\right)\mathclose{}\right\rvert\mathclose{}= 0 \)$ by Proposition II.40 — so $\( f \)$ is bounded on $\( \mathbb{C} \)$. Then, by Liouville's Theorem, $\( f \)$ must be constant — and so $\( f\equiv 0 \)$. This means for all $\( λ\in \mathbb{C} \)$ and for all $\( φ\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{}^{*} \)$, we have $\( φ\mathopen{}\left( {\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}\right)\mathclose{}= 0 \)$.

It follows from the Hahn-Banach Theorem that $\( {\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}= 0 \)$ for every $\( λ \)$, which is impossible.