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
The facts you state do not depend on $K$ being compact, but only on $K$ being selfadjoint. For example, suppose $K^2x=0$. Then $0=\langle K^2x,x\rangle=\langle Kx,Kx\rangle=\|Kx\|^2$ implies $Kx=0$. So, $$ \mathcal{N}(K^2)=\mathcal{N}(K). $$ Therefore, if $K^2=K^3$, then $K^2(I-K)=0$ implies $K(I-K)=0$ or $K^2=K$.
There are several ways to prove that either $\|K\|$ or $-\|K\|$ is in the spectrum of $K$. One way is the prove that the norm of $K$ is the same as the spectral radius, which can be done by showing that $\|K\|=\|K^2\|^{1/2}$ for any selfadjoint operator $K$. This is because the spectral radius is $r_{\sigma}(K)=\lim_{n}\|K^n\|^{1/n}$, and $$ \|K\|=\|K^2\|^{1/2}=\|K^4\|^{1/4}=\cdots=\lim_{n}\|K^{2^n}\|^{1/2^n}=r_{\sigma}(K). $$ To see that $\|K\|=\|K^2\|^{1/2}$ for any selfadjoint $K$, note that $$ \|K^2\| \le \|K\|\|K\|=\|K\|^2, $$ and $$ \|K\|^2=\sup_{\|x\|=1}\|Kx\|^2 = \sup_{\|x\|=1}(K^2x,x) \le \sup_{\|x\|=1}\|K^2x\|\|x\|=\|K^2\|. $$ Every non-zero point of the spectrum $\sigma(K)$ of a selfadjoint compact operator is an eigenvalue. So, either $\|K\|=0$ or at least one of $\|K\|,-\|K\|$ is such a non-zero point of the spectrum because $\|K\|=r_{\sigma}(K)$.
An alternative method for showing that either $\|K\|$ or $-\|K\|$ is in the spectrum relies on the following operator norm equality which holds for selfadjoint operators $K$: $$ \sup_{\|x\|} |\langle Kx,x\rangle| = \|K\|. $$ Either $\sup_{\|x\|=1}\langle Kx,x\rangle =\|K\|$ or $\inf_{\|x\|=1}\langle Kx,x\rangle =-\|K\|$. The first case may be assumed to hold by replacing $K$ with $-K$ if necessary. Let $\{ x_n \}$ be a sequence of unit vectors chosen so that $\langle Kx_n,x_n\rangle$ converges to $\|K\|$. Then $$ \|\|K\|x_n-Kx_n\|^2=\|K\|^2-2\|K\|\langle Kx_n,x_n\rangle+\langle Kx_n,x_n\rangle^2\rightarrow 0. $$ Because $K$ is compact, there is a subsequence $\{ x_{n_k} \}$ such that $\{ Kx_{n_k}\}$ converges to some $y\in \mathcal{H}$. Then $\{ x_{n_k} \}$ also converges: $$ x_{n_k} = \frac{1}{\|K\|}(\|K\|x_{n_k}-Kx_{n_k})+\frac{1}{\|K\|}Kx_{n_k}\rightarrow 0+\frac{1}{\|K\|}y. $$ Hence $\frac{1}{\|K\|}y$ is a unit vector and, by the continuity of $K$, one obtains $K\left(\frac{1}{\|K\|}y\right)=y$, which proves that $\|K\|$ is an eigenvalue.