Lemma (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 been able to prove this with the continuous functional calculus, as I show below (is this proof correct?) but I wondered if there's a way to show it without functional calculus.
Here is one proof of lemma 1 using the continuous functional calculus: Proof. As $A = A^*$ we have $\sigma(A) \subset \mathbb R$, implying $\lambda \not\in \sigma(A)$. Therefore, \begin{equation*} f: \sigma(A) \to \mathbb C, \ x \mapsto \frac{1}{x - \lambda} \end{equation*} is continuous and by the continuous functional calculus we have \begin{align*} \| (A - \lambda I)^{-1} \| & = \| f(A) \| = \sup_{\mu \in \sigma(A)} | f(\mu) | = \sup_{\mu \in \sigma(A)} \frac{1}{| \mu - \lambda |} \\ & = \frac{1}{\inf_{\mu \in \sigma(A)} | \mu - \lambda |} = \frac{1}{\inf_{\mu \in \sigma(A)} \sqrt{(\mu - \Re(\lambda))^2 + \Im(\lambda)^2}} \\ & \le \frac{1}{\sqrt{\Im(\lambda)^2}} = \frac{1}{| \Im(\lambda) |}, \end{align*} where equality holds if $\Re(\lambda) \in \sigma(A)$. $\square$
If $A=A^*$, then $\langle Ax,x\rangle$ is real for all $x\in H$. So $$ -\Im\lambda\|x\|^2=\Im \langle (A-\lambda I)x,x\rangle \\ |\Im\lambda|\|x\|^2 \le \|(A-\lambda I)x\|\|x\| \\ |\Im\lambda|\|x\| \le \|(A-\lambda I)x\| \\ |\Im\lambda|\|(A-\lambda I)^{-1}x\| \le \|x\| \\ \|(A-\lambda I)^{-1}x\|\le \frac{1}{|\Im\lambda|}\|x\| \\ \|(A-\lambda I)^{-1}\|\le\frac{1}{|\Im\lambda|} $$