The question has already been answered above (by Mike Jury). Here is another way of arguing: Any operator with countable spectrum is in the closure of the invertible operators. So are their limits. But again, the unilateral shift provides an example of an operator not in the closure of the invertibles. For if $\|S-X\|<1$ then $\|I-S^*X\|<1$, and so $S^*X$ is invertible. $X$ cannot be invertible, since otherwise $S^*$ would also be invertible (which is). So no translate $S+\lambda I$ can be approximated by operators with countable spectrum.

The question has already been answered above (by Mike Jury). Here is another way of arguing: Any operator with countable spectrum is in the closure of the invertible operators. So are their limits. But again, the unilateral shift provides an example of an operator not in the closure of the invertibles. For if $\|S-X\|<1$ then $\|I-S^*X\|<1$, and so $S^*X$ is invertible. $X$ cannot be invertible, since otherwise $S^*$ would also be invertible. So no translate $S+\lambda I$ can be approximated by operators with countable spectrum.