Yes, this is always true if $\lambda \not= 0$. More generally, the following holds:

**Proposition.** Let $\lambda \not= 0$ be a semi-simple eigenvalue of the compact operator $T$, which means that $\ker((\lambda - T)^2) = \ker(\lambda- T) \not= \{0\}$. Then the eigenspaces $\ker(\lambda - T)$ and $\ker(\overline{\lambda} - T^*)$ separate each other. 

*Proof.* Since $T$ is compact and $\lambda \not= 0$, the generalized eigenspace of $\lambda$ is the range of the corresponding spectral projection $P$. And since $\lambda$ is semi-simple, the generalized eigenspace coincides with the eigenspace $\ker(\lambda-T)$. Hence, $PH = \ker(\lambda - T)$. By applying the same arguments to the dual operator $T^*$ we see that the range $P^*H$ of $P^*$ coincides with $\ker(\overline{\lambda}-T^*)$. But since $P$ is a projection, the ranges of $P$ and $P^*$ separate each other. $\square$

Note that this observation can easily be generalized to Banach spaces (with dual operators rather than adjoints, and the dual eigenvalue $\lambda$ rather than $\overline{\lambda}$).