Set of all subsets of X that contain a set Q is a topology

Question

Let $X$ be a set such that $Q \subset X$.
Show that $\tau = \{\emptyset\} \cup \{U \in \mathcal{P}(X): Q \subset U\}$ is a topology on X.

$\emptyset \in \tau$ by definition and $X \in \tau$ because it contains $Q$.

Let $U_1, U_2 \in \tau$. Thus, ($U_1 = \emptyset$ or $Q \subset U_1$) and ($U_2 = \emptyset$ or $Q \subset U_2$).
If $ \;U_1 = \emptyset$ or $U_2 = \emptyset$, then $U_1 \cap U_2 = \emptyset \in \tau$.
If $\;Q \subset U_1 $ and $Q \subset U_2$, Then $Q\subset (U_1 \cap U_2)$. Thus, $\tau$ is closed under finite intersection.

Let $\mathscr{C} \subset \tau$. Thus for any open set $U \in \mathscr{C}, U = \emptyset$ or $Q \subset U$.
If every open set $U$ in $\mathscr{C}$ is empty, then $\cup\mathscr{C} = \emptyset \in \tau$. Thus, assume that at least one open set $U'$ in $\mathscr{C}$ is nonempty. Thus, $U'$ contains $Q$. Since $U'$ $\subset \cup \mathscr{C}$, $Q \subset \cup \mathscr{C}$. Thus, $\cup \mathscr{C} \in \tau$.

I had originally messed up the last part by not considering the separate case when all open sets in the collection were empty.

Did I miss any important or minor details? Any alternative ways of proving closure under finite intersection or closure under arbitrary union that might be useful to know in general?

Answer 1

$\emptyset \in \tau$ by definition and $X \in \tau$ because it contains $Q$.

Let $U_1, U_2 \in \tau$. Thus, ($U_1 = \emptyset$ or $Q \subset U_1$) and ($U_2 = \emptyset$ or $Q \subset U_2$).
If $ \;U_1 = \emptyset$ or $U_2 = \emptyset$, then $U_1 \cap U_2 = \emptyset \in \tau$.
If $\;Q \subset U_1 $ and $Q \subset U_2$, Then $Q\subset (U_1 \cap U_2)$. Thus, $\tau$ is closed under finite intersection.

Let $\mathscr{C} \subset \tau$. Thus for any open set $U \in \mathscr{C}, U = \emptyset$ or $Q \subset U$.
If every open set $U$ in $\mathscr{C}$ is empty, then $\cup\mathscr{C} = \emptyset \in \tau$.
Now consider the case where at least one open set $U'$ in $\mathscr{C}$ is nonempty.
Thus, $U'$ contains $Q$. Since $U'$ $\subset \cup \mathscr{C}$, $Q \subset \cup \mathscr{C}$. Thus, $\cup \mathscr{C} \in \tau$.