Suppose $\dim X>0$ and $k$ is algebraically closed and uncountable. Moreover, if a "variety" is not necessarily irreducible, the $Z_i$ are supposed to have positive codimension in $X$ (otherwise one could take the irreducible components of $X$).
As in MP's answer, one can suppose $X$ is affine. By Noether's Normalization Lemma, there exists a finite surjective morphism $p: X\to \mathbb A^m_k$ with $m=\dim X$. Let $Y_i=p(Z_i)$. This is a closed subset of $\mathbb A^m_k$ of positive codimension. Moreover $\mathbb A^m_k(k)=\cup Y_i(k)$ because $k$ is algebraically closed (which implies that $Y_i(k)=p(Z_i(k))$). As $k$ is uncountable, there exists a hyperplane $H$ in $\mathbb A^m$ not contained in any $Y_i$ (note that $H\subseteq Y_i$ is equivalent to $H=Y_i$). So by induction on $m$ we are reduced to the case $m=1$, and the assertion is obvious.
Without the hypothesis $k$ algebraically closed, one can show similarly that $X\ne \cup_i Z_i$. This is Exercise 2.5.10 in my book. EDIT In fact this statement is trivial because the generic points of $X$ don't belong to any of the $Z_i$'s. But the proof shows that the set of closed points of $X$ is not contained in $\cup_i Z_i$.