Consider a set $X$ with topologies $\tau_1$ and $\tau_2$. Let $\tau=\tau_1\lor\tau_2$ be the collection of subsets $W\subseteq X$ such that for every $x \in W$ there are subsets $U,V \subset X$ with $x \in U$, $x\in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$. This question has three parts.
1.) Show that $\tau$ is a topology on $X$.
For this part, I know the properties that I have to show. It is clear that both $\emptyset \in \tau$ and $X \in \tau$. I know that I need to show that finite intersections and arbitrary unions remain in $\tau$. To start, I let $W_1,W_2, . . ., W_n \in \tau$. If $W_i = \emptyset$ for any $1 \leq i \leq n$, then the intersection is $\emptyset$, and the union is unaffected, so we can assume that each $W_i$ is nonempty. I don't know where to go from here.
2.) Show that both $\tau_1 \subseteq \tau$ and $\tau_2 \subseteq \tau$.
This seems to me to be trivial, so I am afraid that I might be missing something here. Some guidance in the right direction other than a direct answer would be appreciated.
3.) If $\tau'$ is a topology on $X$ such that $\tau_1,\tau_2 \subseteq \tau'$, then $\tau \subseteq \tau'$.
Let $W \subseteq \tau$. Then, for every $x \in W$ there are subsets $U,V \subset X$ with $x \in U$, $x\in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$. But then since $\tau_1,\tau_2 \subseteq \tau'$, we have $U,V \in \tau'$, and thus $U \cap V \subseteq \tau'$. I'm not sure how to jump from here to saying that $W \subseteq \tau'$.