In my last question on this site (eigenvalues of operator with strange commutator), I found a matrix $A+B$ with the commutator
$$[A,B]=\lambda (A+B)$$
Where $A_{nm}=kn(\delta_{nm}+\delta_{n,m+1})$ and $B_{nm}=(kn-\lambda)(\delta_{nm}+\delta_{n,m-1})$ for any constant $k$, and $n,m\in\mathbb N$. This commutator can be used to show that if $\lambda$ and $A+B$ are nonzero, then $A+B$ must have a continuum of eigenvalues (either half or the entire real number line). I proved this in an answer to my own question linked above.
Because of how it is defined, $A+B$ acts on a vector space with a countable basis, but somehow it has an uncountable number of eigenvalues. How is this possible?