Let $\mathcal{H}$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$. An operator $T\in\mathcal{B}(\mathcal{H})$ is called self adjoint, if $\langle Tx,y\rangle=\langle x, Ty\rangle$ for all $x,y\in\mathcal{H}$.
(a) Show: If $T$ is self adjoint, then $\langle Tx,x\rangle\in\mathbb{R}$ for each $x\in\mathcal{H}$.
(b) Show: If $T$ is self adjoint, then $T$ is invertible if and only if $T$ is bounded below.
(c) Show: If $T$ is self adjoint, then $\|(T\pm i)x\|^2\ge\|x\|^2$ for each $x\in\mathcal{H}$.
(d) Show: If $T$ is self adjoint, then $\sigma(T)\subseteq[-\|T\|, \|T\|]\subseteq\mathbb{R}$.
I have shown (a) and (b) but I am struggling more with (c) and (d) and I want to use (b) and (c) to prove (d). I have seen some other proofs for the fact that $\sigma(T)\subseteq\mathbb{R}$ but I was wondering if someone wouldn't mind illustrating the proof for (d) via (b) and (c), any help on (c) or (d) is greatly appreciated! The work that I have so far is below
(a) If $T$ is self adjoint then \begin{align*} \langle Tx,x\rangle=\langle x,Tx\rangle=\overline{\langle Tx,x\rangle}, \end{align*} hence $\langle Tx,x\rangle\in\mathbb{R}$ for each $x\in\mathcal{H}$.
(b) $\implies$ Suppose that $T$ is invertible, then $T^{-1}Tx=Ix=x$ for all $x\in\mathcal{H}$. Thus, for all $x$ we have \begin{gather*} \|x\|=\|T^{-1}Tx\|\le\|T^{-1}\|\|Tx\|\\ \implies \|Tx\|\ge\frac{1}{\|T^{-1}\|}\|x\|, \end{gather*} hence $T$ is bounded below by definition.
$\Longleftarrow$ Suppose that $T$ is bounded below, then there exists some $C>0$ such that \begin{align*} \|Tx\|\ge C\|x\|\,\,\text{for all $x\in\mathcal{H}$}. \end{align*} Thus, if $Tx=0$, then $0\ge C\|x\|\implies\|x\|=0\implies x=0$. Hence, $T$ is injective. Now, we also have that \begin{gather*} \text{range}(T)^\perp=\text{range}(T^*)^\perp=\text{ker}(T)=\{0\}\\ \implies\overline{\text{range}(T)}=\text{range}(T)^{\perp\perp}=\mathcal{H}. \end{gather*} Next, we show that $\text{range}(T)$ is closed. To this end, suppose $\{Tx_n\}_{n\ge1}$ is a sequence from range($T$) which converges to some $y\in\mathcal{H}$. Since $\{Tx_n\}$ is a convergent sequence, it is Cauchy and thus $\{x_n\}$ must be a Cauchy sequence as well since $\|Tx_m-Tx_n\|\ge C\|x_m-x_n\|$ for all $m,n\in\mathbb{N}$. Therefore, $\{x_n\}$ converges to some $x\in\mathcal{H}$, hence \begin{align*} Tx=\lim_{n\to\infty}Tx_n=y. \end{align*} Thus, $y\in\text{range}(T)$ and so range($T$) is closed. Therefore, $\text{range}(T)=\mathcal{H}$ and so $T$ is also surjective, hence $T$ is invertible.
For c, note that $$ \langle (T\pm i)x,(T\pm i)x\rangle=\langle x,(T\mp i)(T\pm i)x\rangle= \langle x,(T^2+I)x\rangle=\langle x,T^2x\rangle+\langle x,x\rangle $$ which is clearly bounded below by $||x||^2$. Note how this can be easily modified to show that $||(A-z)x||\geq |\Im(z)|||x||^2$.
To conclude, d) you can pretty much repeat your reasoning in part b to show that $T-z$ has a bounded inverse whenever $\Im(z)\ne 0$, so that the resolvent exists off the real line.