We know that any compact symmetric operator on a Hilbert space, has a orthogonal base of eigenvectors. But we also know that $0$ is in the spectrum if $X$ is infinite dimensional, which makes the operator non-invertible(in particular they are never surjective).
Why we would care to diagonalise in infinite dimensional at all since never get as good info as in finite dimension (non-zero eigenvalues on the diagonal).
Added later ;
Plausibel answer or claim ;
Diagonalisation of a compact operator $C$ in infinite dimensions seems more related to the invertabiity(or solvability) of Iλ−C (a fredholm equation of 2nd kind) for compact operator C rather then properties of C itself as a linear map which often is the object of intrest in finite dimension i.e linear algebra. That might be a reason for my confusion. From a historcial point of view we seem to care about solvability of Iλ−C and qualitive properties of solutions rather then C itself as a map from one vector space into another,in some sense. So when we "diagonalize" we really just cheack solvabilty of Iλ−C, we are not looking for a nice bijection(the full diagonal matrix) of the vectorspaces as we do in linear algebra.
Or simply the MAIN reason we started to study the spetrum of an operator $C$ was to determine if $I \lambda - C$ is solvable.
Am I on to something or just way off?