It looks like the accepted answer to this question didn't address some of the questions asked in the comments above, so I thought I'd fill in some details. This is really just a long comment; sorry I had to include it as an answer.
It seems that the question is regarding the Zariski topology on two-sided prime ideals. The word "ideal" below refers to two-sided ideals.
A prime idealprime ideal of a noncommutative ring $R$ is a proper ideal such that, for any ideals $I$ and $J$ of $R$, if the ideal productideal product $IJ$ is contained in $P$, then either $I \subseteq P$ or $J \subseteq P$. Let $\mathrm{Spec}(R)$ denote the set of prime ideals of $R$.
The Zariski topology on $\mathrm{Spec}(R)$ can be defined just as in the commutative casecommutative case, as stated in Simon Rose's answer: the closed subsets of $\mathrm{Spec}(R)$ are precisely those of the form $V(I) := \{P \in \mathrm{Spec}(R) : I \subseteq P\}$. It's easy to see that $\bigcap_j V(I_j) = V(\sum I_j)$ for any set of ideals $\{I_j\}$ of $R$, where $\sum I_j$ is the smallest ideal of $R$ containing the $I_j$ (and is equal to the set of all finite sums of elements from the $I_j$). To see that these are closed under finite unions, one uses the definition of prime ideal to see that $V(I) \cup V(J) = V(IJ)$. That is, for any prime $P$ of $R$ and any ideals $I$ and $J$ of $R$, the statement $IJ \subseteq P$ is equivalent to the statement that either $I \subseteq P$ or $J \subseteq P$.
