Union of Topologies is not necessarily Topology

Theorem
Let $\left({\tau_i}\right)_{i \in I}$ be an arbitrary indexed set of topologies for a set $X$.

Then $\tau := \displaystyle \bigcup_{i \in I} {\tau_i}$ is not necessarily also a topology for $X$.

Proof
Let $X := \left\{{0, 1, 2}\right\}$ be a set.

Let: be topologies on $X$.
 * $\tau_1 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{0, 1}\right\}, X}\right\}$
 * $\tau_2 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{2}\right\}, \left\{{0, 2}\right\}, X}\right\}$

Then:
 * $\tau := \tau_1 \cup \tau_2 = \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{2}\right\}, \left\{{0, 1}\right\}, \left\{{0, 2}\right\}, X}\right\}$

For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$.

But:
 * $\left\{{1}\right\} \cup \left\{{2}\right\} = \left\{{1, 2}\right\} \not \in \tau$

Therefore $\tau$ is not a topology on $X$.

Hence the result.

Remark
Since the result does not hold for these two particular topologies, it certainly does not hold in general.

However if $\left|{X}\right| \le 2$ and $\left({\tau_i}\right)_{i \in I}$ is an arbitrary indexed set of topologies for $X$, then $\tau := \displaystyle \bigcup_{i \in I} {\tau_i}$ is also a topology for $X$.

See: Union of Topologies on Singleton or Doubleton is Topology for this result.