If $\{\tau_\alpha\}$ is a family of topologies on $X$, show that $\cap \tau_\alpha$ is a topology on $X$. Is $\cup \tau_\alpha$ a topology on $X$?
For all $\alpha$, $\varnothing \in \tau_\alpha$ and $X \in \tau_\alpha$. So, $\varnothing \in \cap \{\tau_\alpha\}$ and $X \in \cap \{\tau_\alpha\}$. Now, let $\{U_\beta\}_{\beta \in J}$ be an indexed collection of subsets of $\cap \{\tau_\alpha\}$. Then, for all $\alpha$, $\{U_\beta\}_{\beta \in J} \subseteq \tau_\alpha$. This implies that for all $\alpha$, $\cup\{U_\beta\}_{\beta \in J}\in\tau_\alpha$, since $\tau_\alpha$ is a topology. But then, $\cup\{U_\beta\}_{\beta \in J}\in\cap \{\tau_\alpha\}$.
Now, let $U_1, U_2, \ldots U_n$ be subsets of $\cap \{\tau_\alpha\}$. Then, for all $\alpha$, $U_i \in \tau_\alpha$. This implies that for all $\alpha$, $\cap_{i=1}^n U_i \in \tau_\alpha$, since each $\tau_\alpha$ is a topology. But then, $\cap_{i=1}^n U_i \in \cap \{\tau_\alpha\}$. Therefore, $\cap \{\tau_\alpha\}$ is a topology on $X$.
However, it is not true that $\cup \{\tau_\alpha\}$ is a topology. Consider the set $X = \{a,b,c\}$ and the topologies $\tau_1 = \{\varnothing, X, \{a\}\}$ and $\tau_2 = \{\varnothing, X, \{b\}\}$. It can clearly be seen that $\tau_1 \cup \tau_2$ does not define a topology on $X$.
Best Answer
Looks good to me. Also, your counterexample is the minimal one (i.e. if $|X|$ was smaller, there exits no such counterexample).