This is true in general; you do not need $X$ to be a noetherian space.
Continuing user3296's argument: the only thing left to check is that $C \cap Z$ and $D \cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.
To see that for example $C \cap Z \subsetneq Z$, you have to convince yourself that $C \cap Z = Z$ implies $C = \overline{C} = \overline{C \cap Z} = \overline Z$. The first equality is true because $C$ is closed in the closed set $\overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = \overline C$ is contained in $\overline Z = \overline{C \cap Z}$, and on the other hand, $C$ contains $C \cap Z$ so that $\overline C$ contains $\overline{C \cap Z}$.