I think I have it. Take $x = Ay / \|Ay\|$, then
$$
\|Ay\| = |\frac{\|Ay\|^2}{\|Ay\|}| = |\langle x,Ay\rangle |
$$
so the inequality follows once we know that
$$
|\langle x, Ay\rangle| \le \sup_{\|x\| = 1} \langle x,Ax \rangle
$$
and the latter was given.
Self-adjointness is used to show that $\langle Ax,x\rangle$ is always real. Recall that $\langle u,v\rangle = \overline{\langle v,u\rangle} $ and use the definition of self-adjointness:
$$
\langle Ax,x\rangle=\langle x,Ax\rangle = \overline{\langle Ax,x\rangle}
$$
The boundedness of $A$ implies that the set $\{\langle Ax,x\rangle:\|x\|=1\}$ is bounded. So, it's a bounded subset of $\mathbb R$. Now $m$ and $M$ make sense and the proof goes as before.
By the way, there is a gap in is bounded below and is hence invertible: for example, the shift operator $(x_1,x_2,\dots)\mapsto (0,x_1,x_2,\dots)$ is bounded from below but is not invertible. This is another place where self-adjointness will be invoked: re-read the proof to see what goes on there.
If you don't need bounds for the spectrum, you could simplify the proof by dropping $M$ and $m$. Since the modulus of a complex number is at least the modulus of its imaginary part,
$$
\|(A- \lambda I)x\| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge |\operatorname{Im} \lambda|
$$
Best Answer
It is clear that $|\langle Tx,x\rangle|\leq \|T\|$ for $\|x\|=1$. For the converse, it suffices to show that $|\langle Tx,y\rangle|\leq \alpha$ for all $\|x\|=\|y\|=1$, with $\alpha=\sup\bigl\{|\langle Tx,x\rangle|: \|x\|=1\bigr\}$. We can clearly assume $\langle Tx,y\rangle \in\mathbb R$. Then $$ \langle Tx,y\rangle = (\langle T(x+y),x+y\rangle - \langle T(x-y),x-y\rangle)/4. $$ But then $$ |\langle Tx,y\rangle|\leq\alpha(\|x+y\|^2+\|x-y\|^2)/4=\alpha, $$ by the parallelogram identity.
I have paraphrased this nice derivation from the book Essential Results of Functional Analysis, by Zimmer.