The spectrum of an operator is often uncountable, but please note that the question refers to eigenvalues.
This is well known for self-adjoint operators, but I don't see how the result extends beyond that case.
The same question also makes sense for separable Banach spaces.
The same question is asked and answered on mathoverflow here. The backwards shift operator $T : \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ defined by $$T(a_0,a_1,a_2,\ldots) = (a_1,a_2,a_3,\ldots)$$ does the job. For every $\lambda \in \mathbb{C}$ with $|\lambda|<1$, the geometric sequence $v = (1,\lambda,\lambda^2,\ldots)$ is square-integrable and satisfies $T v = \lambda v$.