Let's consider the following inclusion chain of topologies on space $X$: $\tau_1\subset\tau_2\subset\cdots\subset\tau_n\subset\cdots$. Let $\tau=\bigcup_{n=1}^\infty \tau_n$. Is $\tau$ a topology?
Obviously , the intersection of any two sets from $\tau$ belongs to $\tau$. However, it is not clear whether $\bigcup_{n=1}^\infty A_n\in\tau$ where $A_n\in\tau_n$. I think, in general $\tau$ is not a topology but cannot find a counterexample.
Indeed it is not; the trick is to put new open sets in each $A_n$ so that you can take the union of one from each $A_i$ which is not in any. For example, take $X = \mathbb R$, and let $\tau_n$ consist of $\mathbb R$ itself, together with all sets which are open in the Euclidean topology and which are within the interval $(-n, n)$. Then for each $k \in \mathbb N$, the open set $(k, k + 1)$ is in some $\tau_i$, yet $$ \bigcup_{k \in \mathbb N} (k, k+1) = \mathbb R_{>0} \setminus \mathbb N $$ does not lie in any $\tau_n$.
In a similar spirit to Mees de Vries answer, let $\tau$ be the standard topology on the rational numbers $\mathbb{Q} = \{q_1,q_2,q_3,\ldots\}$. Define $\tau_n$ to be the coarsest topology refining $\tau$ in which each of the points $q_1,\ldots,q_n$ is open. If you want a way to picture this topology geometrically, note that the map $f_n : \mathbb{Q} \to \mathbb{R}^2$ given as $$f_n(x) = \begin{cases} (x,0) & \text{ if } x \notin \{q_1,\ldots,q_n\} \\ (x,1) & \text{ if } x \in \{q_1,\ldots,q_n\} \\ \end{cases}$$ is homeomorphism from $(\mathbb{Q},\tau_n)$ onto $f(\mathbb{Q}) \subseteq \mathbb{R}^2$.
If $\bigcup_{n=1} \tau_n$ was a topology on $\mathbb{Q}$, it would have to be the discrete topology. But this is not the case because, for example, $\mathbb{Z} \subseteq \mathbb{Q}$ does not belong to any of the topologies $\tau_n$.
Not usually. Let $X=\mathbb N$ and $\tau_n=\{\phi, X, \{n\}\}.$ Then $\cup \{\tau_n:n\in \mathbb N\}=\{\phi, X\}\cup \{\{n\} :n\in \mathbb N\}$ is not a topology because it contains $\{1\}$ and $\{2\}$ but does not contain $\{1\}\cup \{2\}=\{1,2\}.$
Even if $\tau_n\subset \tau_{n+1}$ for all $n,$ this may fail.
Example:Let $X=\mathbb N.$ For $n\in \mathbb N,$ let the members of $\tau_n$ other than $X$ be all of the subsets of $\{m\in \mathbb N:m\leq n\}.$ The members of $T=\cup \{\tau_n: n\in \mathbb N\},$ other than $X,$ are all of the finite subsets of $X,$ and $T$ is not a topology. $T$ contains $\{n\}$ for every $n\in X,$ so if $T$ $were$ a topology then every subset of $X$ would belong to $T.$ But $X\supset \{2n:n\in \mathbb N\}\not \in T.$
