How would I write a proof for this example?
Let X be a set, and let $B \subseteq \mathcal P \left({X}\right)$. Define
$B^* =${ $U \subseteq X:$ There is an index set $I$ and $U_{i} \in B$ for each $i \in I$ with $U = \bigcup\limits_{i\in I} U_{i}$ }.
Informally, we think of $B^*$ as "the closure of $B$ under arbitrary unions."
If $B \subseteq \mathcal P \left({X}\right)$, we say that $B$ is a basis on X, if for any $U, V \in B$ and any $x \in U \cap V$, there is $W \in B$ with $x \in W$ and $W \subseteq U \cap V.$
Let X be a set and $B$ a basis on X.
Prove by induction on $k$ that if $U_{0},...,U_{k-1} \in B$ and $U_{0} \cap...\cap U_{k-1} \in B$, then there is $W \in B$ with $x \in B$ and $W \subseteq U_{0} \cap...\cap U_{k-1}$.
Then, prove that $B^*$ is a topology on X.
The induction on $k$ is standard: You can start with $k=1$ (we then have one set and the statement is trivially true). Assume truth for $k \ge 1$. For $k+1$, write $U_0 \cap \ldots \cap U_{(k+1)-1}$ as $(U_0 \cap \ldots U_{k-1}) \cap U_k$, and apply the hypothesis for the left hand side, and the given base property for the conclusion.
To check we have a topology, show that the property from the induction shows that $B^\ast$ is closed under finite intersections. That it is closed under unions is true by construction almost; Taking the index set $I = \emptyset$ ensures that $\emptyset \in B^\ast$.
I don't see how $X \in B^\ast$ in general, as we could take $B = \emptyset$, and then $B^\ast = \{\emptyset\}$ is not a topology. We need to assume that $\cup B = X$ as well.