Is it true that every closed operator on a separable Hilbert H space only has countably many eigenvalues?

Or put the other way around, if I want to ensure that a (not necessarily bounded) linear operator on a separable Hilbert space only has countably many eigenvalues, is closedness (or better said, closability) a sufficient condition?

(By the term eigenvalue, I do not only mean a point in the spectrum of course, but one that actually fulfills $Tx = \lambda x$.)