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,
$$
\|(A- \lambda I)x\| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge |\operatorname{Im} \lambda|
$$
Assume $T$ has empty spectrum. Then $T$ is invertible, $T^{-1}$ is a bounded selfadjoint operator and, for $\lambda \ne 0$,
$$
(T^{-1}-\lambda I) =(I-\lambda T)T^{-1}=\lambda(\frac{1}{\lambda}I-T)T^{-1}
$$
has bounded inverse
$$
\frac{1}{\lambda}T\left(\frac{1}{\lambda}I-T\right)^{-1}
$$
So $\sigma(T^{-1})=\{0\}$ because only $\lambda=0$ can be in the spectrum, and it cannot be empty. But that implies $T^{-1}=0$, which is a contradiction.
Best Answer
Note that (b) implies that the range of $B$ is dense. If we can show that it's all of $H$, then we'll know that $B$ is bijective, which means that $\lambda\notin\sigma(A).$ This will follow from (c). Indeed, take $x\in H.$ By density, there exists a sequence $(y_n)$ in $H$ so $By_n\rightarrow x.$ Using part (c), we get that $(y_n)$ is Cauchy, so it converges, say to $y\in H$. By continuity, we get that $By=x$, so $x\in B(H).$ The other inclusion is obvious.
All that's left is to show part (c). Simply observe that
\begin{align*} \|x_n-x_m\|^2&=\|By_n-By_m\|^2=\|(\lambda-A)(y_n-y_m)\|^2\\ &=\|(\text{Re}\ \lambda-A)(y_n-y_m)+i\text{Im}\ \lambda(y_n-y_m)\|^2\\ &= \|(\text{Re}\ \lambda-A)(y_n-y_m)\|^2+\|\text{Im}\ \lambda(y_n-y_m)\|^2\\ &\geq |\text{Im}\ \lambda|^2\|y_n-y_m\|^2 \end{align*} (Self-adjointness was used when moving from the second to the third line.)