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 $\sigma(A)\subseteq\mathbb{R}$.
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
You can't deduce that $\lambda = \overline{\lambda}$ immediately. All you know is that if $\lambda \in \sigma(T)$, then there is $\mu \in \sigma(T)$ with $\lambda = \overline{\mu}$. How do you conclude $\lambda = \overline{\lambda}$ from this?