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,
$$
\|(A- \lambda I)x\| \ge |\langle (A- \lambda I)x, x \rangle| =|\langle Ax, x \rangle - \lambda| \ge |\operatorname{Im} \lambda|
$$
Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.
$$(Tf)(x) = x\cdot f(x).$$
Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.
Clearly $T$ has no eigenvalues, since
$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$
and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.
We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.
The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.
Best Answer
If $\Im\lambda \ne 0$, and $x \in X$, then $$ \Im\lambda \|x\|^{2} = -\Im((A-\lambda I)x,x),\\ |\Im\lambda|\|x\|^{2} \le |((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\|,\\ |\Im\lambda|\|x\| \le \|(A-\lambda I)x\|. $$ So $A-\lambda I$ is injective for all $\lambda\notin\mathbb{R}$. The above inequality can be used to show that the range $\mathcal{R}(A-\lambda I)$ is closed for $\lambda\notin\mathbb{R}$. So $A-\lambda I$ is surjective for $\lambda\notin\mathbb{R}$ because $$ \begin{align} \mathcal{R}(A-\lambda I)& =\overline{\mathcal{R}(A-\lambda I)} \\ & =\mathcal{N}(A^{\star}-\overline{\lambda}I)^{\perp} \\ & = \mathcal{N}(A-\overline{\lambda}I)^{\perp}=\{0\}^{\perp}=H. \end{align} $$ Therefore, $A-\lambda I$ is injective and surjective for $\lambda\notin\mathbb{R}$, which leaves $\sigma(A)\subseteq\mathbb{R}$.