I am thoroughly confused at the moment. I have a result which shows that for compact and symmetric operators $K$ from the Hilbert space $H$ to itself, $\pm||K||$ is an eigenvalue.
But without requiring compactness on $K$, I can get the following assuming I have eigenvalues to begin with such that $Kx=\lambda x$.
$\lvert\langle Kx,x\rangle\rvert = \lvert\langle\lambda x,x\rangle\rvert = \lvert\lambda\rvert \lVert x\rVert^2$
Taking the supremum,
$\sup\{\lvert\langle Kx,x\rangle\rvert,\ \lVert x\rVert=1\} = \lvert\lambda\rvert = \lVert K\rVert$
What is wrong with my reasoning here?
Not every set of reals has a supremum, only bounded non-empty sets do. You are making two unstated assumptions: the set $\{\lvert\langle Kx,x\rangle\rvert:x\text{ an eigenvalue}\}$ is bounded, and that it is non-empty. Without the first assumption, $\sup\lvert\langle Kx,x\rangle\rvert$ may not exist, (or be $\infty$) if the set is unbounded. Assuming it exists is equivalent to assuming your operator $K$ is bounded.
So should we conclude that for all bounded $K$, $\lVert K\rVert$ is an eigenvalue? No, we're not there yet. We need also that $\{\lvert\langle Kx,x\rangle\rvert:x\text{ an eigenvalue}\}$ is nonempty.
For bounded symmetric operators it is true that $\lVert K\rVert=\sup_{\lambda\in\sigma(K)}\lvert \lambda\rvert.$ The norm of the operator is the largest value in the spectrum, the spectral radius. Recall that the spectrum $\sigma(K)$ of an operator is the set of points such that $K-\lambda$ is not invertible. In finite dimensions, an operator is singular iff it has a nontrivial nullspace. In other words, a point is in the spectrum iff it is an eigenvalue with an eigenvector. However in infinite dimensions, non-invertibility does not guarantee that $K-\lambda$ has a nullspace. There are other ways to fail to be invertible in infinite dimensions, for example it can fail to be surjective. Or $(K-\lambda)^{-1}$ may exist but fail to be continuous. So in infinite dimensions even a symmetric operator over an algebraically closed field can fail to have eigenvalues.
But compactness of the operator $K$ is a sufficient condition that a non-zero point of the spectrum be an eigenvalue with eigenvectors. Note that it is not a necessary condition. For example the identity operator is not compact, but has eigenvalues and eigenvectors.
For example, let $H=L^2([0,1])$ and $Kf=xf(x)$. Then the norm of the operator is $\lVert K\rVert = 1.$ And the spectrum of $K$ is $[0,1]$. The spectral radius is $\sup[0,1]=1.$ But $1$ is not an eigenvalue, because there are no non-zero $L^2$ functions satisfying $xf(x)=f(x).$ In fact this operator has no eigenvalues. How can it be that $1$ is in the spectrum but is not an eigenvalue? Because $K-1$ is injective (since $K$ has no eigenvectors) but not surjective (since $1/(x-1)$ is not in $L^2$, there is no $f$ such that $(x-1)f(x)=1.$) This operator is not compact. To see this directly, observe that $f_n=\sqrt{n}\cos(nx)$ is a bounded sequence, but $\lVert Kf_n\rVert\to \infty.$ Or consider the shift operator on $\ell^2(\mathbb{N}).$
To guarantee that a point in the spectrum is actually an eigenvalue with an eigenvector, at least for the nonzero spectrum, requires an additional assumption like compactness of the operator. This result, for compact operators, is called the Fredholm alternative: if $K$ is compact an $I$ is an isomorphism, then $K+I$ either has a finite dimensional kernel, or is invertible. Terry Tao has a nice blog post containing some proofs of this result, and discussion.
Note that zero may be in the spectrum of a compact operator without being an eigenvalue. For example $K(e_n)=\frac{1}{n}e_n$ on $\ell^2(\mathbb{N}).$ And zero must be in the spectrum, see Spectrum of a compact operator