Number of conjugacy classes vs irreps for continuous compact group What is the intuition behind that the number of conjugacy classes is not equal to the number of irreps in continuous compact groups. The former is uncountable (as the abelian SO(2) demonstrates), while the latter is countable (per Peter-Weyl theorem).
The context of this question is in my previous post 
How many irreducible representations does SO(2) have?, which has been answered by José Carlos Santos. The aim of the present separate post is that I would like to learn about the intuition behind. In particular, I suspect a strong connection with spectrum of some operators.
 A: Let me reformulate your question in a form that is more easily answerable. The number of irreps is the cardinality of an orthonormal basis of $L^2(G)$, where $G$ is the group under consideration. So, let us reason in terms of $L^2$ spaces.

Definition. Let $\Omega$ be a set and let $d\mu$ be a measure on it. We let $L^2(\Omega)$ denote the vector space of all functions $f\colon \Omega\to \mathbb C$ with the property that
$$\tag{1}
\int_{\Omega}|f(x)|^2\, d\mu <\infty.$$
In the following we consider $\Omega=\{1, 2, \ldots, n\}$, with the "counting measure", which means that (1) reads
$$
\int_{\{1, 2, \ldots, n\}} |f(k)|^2\, d\mu = \sum_{k=1}^n |f(k)|^2, $$
and there is no need to require finiteness, as that's just a finite sum. Since a vector in $\mathbb C^n$ can be interpreted as a function from $\{1, 2, \ldots, n\}$ to $\mathbb C$, we see that $L^2(\{1, 2, \ldots, n\})$ is just a fancier way of writing $\mathbb C^n$.
We also consider $L^2(0, 1)$, in which case the measure is the standard Lebesgue measure on $(0,1 )$.

You have noticed that $\mathbb C^n=L^2(\{1, 2, \ldots, n\})$ admits an orthonormal basis made of $\lvert \{1, 2, \ldots, n\}\rvert$ elements, and this led you to conjecture that $L^2(0, 1)$ should have an orthonormal basis made of $\lvert (0, 1) \rvert$ elements. However, this is false, as orthonormal bases of $L^2(0, 1)$ are countable; one example of such a basis is the trigonometric system $\{e^{2\pi i xk}\ :\ k\in\mathbb Z\}$.
The reason for this mismatch is that the two kind of orthonormal bases are deeply different. In the case of $\mathbb C^n$, we are talking of an algebraic basis, while in the case of $L^2(0,1)$ there are many more elements to take into account. First, elements of $L^2(0,1)$ are not just functions of $(0,1)$ into $\mathbb C$, like in the finite-dimensional case; they must also satisfy quite stringent requirements (being measurable and square-integrable). This rules out a lot of functions, dramatically reducing the cardinality of a possible basis. Moreover, the basis itself is not an algebraic one; it involves finite sums, so it requires the introduction of a topology.

A final note. You ask whether the irreps are always the eigenfunctions of some operator. I think that the answer is true, but I don't know for sure. The keyword to search for is "Casimir operator". For $SO(n)$, this operator is the Laplace-Beltrami on the sphere, which indeed has a discrete spectrum. In the comments, you suggest that this is a general phenomenon. It may well be, but I don't know if it is a general fact that a Casimir operator (whatever that is) is the Laplace-Beltrami for some compact Riemannian manifold.
What I would bet on is: the same compactness mechanism that is responsible for the discreteness of the spectrum of a compact Riemannian manifold is responsible for the discreteness of irreps in Peter-Weyl's theorem (follow the link for more information on such compactness in the Riemannian manifold case). To be sure, one should go through the proof of Peter-Weyl and check this.
Final note: you say "discreteness of spectrum for bounded geometries"; I have interpreted this, in mathematical terms, as "discreteness of the spectrum of a compact Riemannian manifold".
