Lemma 1 (Resolvent bound). Let $H$ be a Hilbert space, $A \in L(H)$ linear, bounded and self-adjoint and $\lambda \in \mathbb C \setminus \mathbb R$. Then $$\| (A - \lambda I)^{-1} \| \le | \Im(\lambda) |^{-1}.$$
I have shown here a way to prove lemma 1 with functional calculus and in this answer to the above mentioned question is way to show this result without functional analysis.
A friend suggested to me the following proof of lemma 1, which heavily relies on the following lemma:
Lemma 2 (Invertibility). Let $E$ and $F$ be Banach spaces and $T \in L(E,F)$ be bijective. Let $T^{-1} \in L(F,E)$ be the inverse of $T$ and $S \in L(E,F)$ such that $$ \| S - T \| < \| T^{-1} \|^{-1} $$ Then $S$ is also invertible.
Proof. Assume $\| (A - \lambda I)^{-1} \| > | \Im(\lambda) |^{-1}$. As $\sigma(A) \ne \emptyset$ (Gelfand-Mazur), let $\mu \in \sigma(A)$. As $A$ is self-adjoint, it's spectrum is real, so $| \lambda - \mu | \ge | \Im(\lambda) |$. Thus $$ \| (A - \mu I) - (\lambda - \mu) I \| = \| A - \lambda I \| < | \Im(\lambda) | = | \lambda - \mu | = \| ((\mu - \lambda) I)^{-1} \|^{-1}. $$ By lemma 2 $A - \mu I$ is invertible, which contradicts $\mu \in \sigma(A)$. $\ \square$
Unfortunately, I don't think that we can conclude $$ \| (A - \lambda I)^{-1} \| > | \Im(\lambda) |^{-1} \implies \| A - \lambda I \| < | \Im(\lambda) |. $$ Is there a way to save this proof idea?