When is the union of topologies a topology? The union of two topologies on some set may or may not be a topology. When is it a topology?
 A: Years late to answer this but:
There is actually a problem in Munkres related to this: prove for a family of topologies $\{\tau_\alpha \}$ on a set $X$ that there is a unique smallest topology containing all of the topologies in the family.
If you set $S = \{ U : U \in \tau_\alpha \text{ for some } \alpha \}$ and consider the topology generated by $S$ (as a subbasis) I think this does the trick.  It also provides a less wordy sufficient and necessary condition to your question.
A: Let $\tau_1$ and $\tau_2$ be two topologies on a point set $X$.  That is, $\tau_1$ and $\tau_2$ are subsets of the powerset of $X$, each indicating which subsets of $X$ should be called open.  Let $\tau = \tau_1 \cup \tau_2$.  In general $\tau$ is not a topology on $X$, but let's see which conditions are satisfied and which could fail:
(1)  $\emptyset \in \tau$.  True since $\emptyset \in \tau_1$ (and $\tau_2$).
(2)  $X \in \tau$.  True since $X \in \tau_1$ (and $\tau_2$).
(3)  Arbitrary unions of open sets are in $\tau$, that is,
$$\bigcup_{\alpha \in \mathscr{I}} U_{\alpha} \in \tau.$$
Not true in general.  At best, I can partition the sets into those in $\tau_1$ and those in $\tau_2$ so that we have:
$$\bigcup_{\alpha \in \mathscr{I}} U_{\alpha} = \bigcup_{\alpha \in \mathscr{I} \cap \tau_1} U_{\alpha} \cup \bigcup_{\alpha \in \mathscr{I} \cap \tau_2} U_{\alpha} = V \cup W,$$
where $V \in \tau_1$ and $W \in \tau_2$ because each $\tau_i$ is a topology.  But here's where we get stuck.  I have no guarantee that $V \cup W$ is in either $\tau_1$ or $\tau_2$, hence I can't place $V \cup W \in \tau$.
(4) Finite intersections of open sets are in $\tau$, that is,
$$\bigcap_{i=1}^n U_i \in \tau.$$
Again, I could partition the sets according to $\tau_1$ and $\tau_2$ so that
$$\bigcap_{i=1}^n U_i \in \tau = V \cap W,$$
for $V \in \tau_1$ and $W \in \tau_2$.  But unless $V \cap W \in \tau_1$ or $\tau_2$, I cannot find $V \cap W \in \tau = \tau_1 \cup \tau_2$.
So this leads to the necessary (and sufficient) conditions:  $\tau = \tau_1 \cup \tau_2$ is a topology if every pairwise union $U \cup V$ and intersection $U \cap V$ of open sets $U \in \tau_1$ and $V \in \tau_2$ lies in either $\tau_1$ or $\tau_2$.  Of course, this does not strike me as a very useful or easy condition to check.
