Let $X$ be a set and suppose $\mathbb{B}$ is a collection of subsets of $X$.
$\mathbb{B}$ is a basis for some topology on $X$ $\textbf{if and only if}$
(1)$\bigcup_{B\in \mathbb{B}}B = X$ (Covering property)
(2) If $B_1, B_2 \in \mathbb{B}$ and $x\in B_1 \cap B_2$ then $\exists$ $B_3 \in \mathbb{B}$ such that $x\in B_3 \subseteq B_1\cap B_2$.
My proof:
$\textbf{Forward direction.}$
Suppose $\mathbb{B}$ is a basis for some topology $\tau$ on $X$. Let $x\in X$. As $X$ is $\tau-open$, it follows that $X$ is the union of some collection of elements in $\mathbb{B}'\subseteq \mathbb{B}$. Therefore $x$ must be in some $B\in \mathbb{B}$, hence the union of elements in $\mathbb{B}$ cover $X$. Thus we have shown that $\mathbb{B}$ satisfies the first condition.
Let $B_1, B_2\in \mathbb{B}$. Suppose $x\in B_1\cap B_2$. As $B_1,B_2$ are open, their intersection must be open and so by Lemma 9.1, there exists $B_3 \in \mathbb{B}$ such that $x \in B_3 \subseteq B_1\cap B_2$. Consequently, the second condition has been verified.
$\textbf{Backwards direction}$
Let $\tau =$ $\{$ All unions of elements in $\mathbb{B}$ $\}$.
We will show that $\tau$ is a topology on $X$. As the union of all elements in $\mathbb{B}$ cover $X$, $X\in \tau$. Further, the elements in the empty collection $\varnothing$ $\subseteq \mathbb{B}$, give a union $\varnothing$. Let $(U_i)_{i\in I}$ be arbitrary collection of elements in $\tau$. By definition of $\tau$, the union must be in $\tau$. Let $V_1,V_2$ be any two sets in $\tau$. So $V_1= \bigcup_{B\in \mathbb{B}'}B$ and $V_2= \bigcup_{A\in \mathbb{B}''}A$, where $\mathbb{B}',\mathbb{B}''$ are subsets of $\mathbb{B}$. Then $V_1 \cap V_2$ $=$ $\bigcup_{B \in \mathbb{B}',A\in \mathbb{B}''} B\cap A$; which is the union of elements in $\mathbb{B}$, so $V_1 \cap V_2 \in \mathbb{\tau}$. Hence $\tau$ is a topology on $X$. Now I must show that $\mathbb{B}$ is a basis for $\tau$. Let $B_1 \in \mathbb{B}$. Then, $B_1= B_1\cup B_1$. Therefore $B_1\in \mathbb{\tau}$. So every element in $\mathbb{B}$ is an element of the topology defined above. Every element of the topology defined above is the union of elements in $\mathbb{B}$, by construction of $\tau$. Therefore, $\mathbb{B}$ is a basis for $\tau$.
Is the proof correct? May I please have feedback?
Note: Lemma 9.1 is:
Let $X$ be a topological space and $\mathbb{B}$ a basis for the topology on $X$. Then $U\subseteq X$ is open $\iff$ $\forall p\in U$ $\exists$ $B\in \mathbb{B}$ such that $p\in B \subseteq U$.
The proof of this lemma is:
If $U\subseteq X$ is open, then $U$ is the union of some collection of elements in $\mathbb{B}$ and so there exists a $B'\in \mathbb{B}$ such that $p\in B'\subseteq U$. The converse follows because it shows that the interior of $U$ is equal to $U$ (Since a basis is a collection of open sets).
Best Answer
The backwards direction is not quite correct. Define (instead of the unions) a set as open like in your lemma 9.1 : $U$ is open iff for all $p \in U$ there is some $B \in \Bbb B$ with $p \in B \subseteq U$.
Then the intersection proof becomes: suppose $U_1, U_2$ are open in this sense, and consider $U_1 \cap U_2$, and let $p \in U_1 \cap U_2$. As $p \in U$, there is some $B_1 \in \Bbb B$ with $p \in B_1 \subseteq U_2$, and similarly an $B_2 \in \Bbb B$ with $p \in B_2 \subseteq U_2$. Now we apply (2), to get $B_3$ with $$p \in B_3 \subseteq B_1 \cap B_2 (\subseteq U_1 \cap U_2)$$ and as $p \in U_1 \cap U_2$ was arbitrary, $U_1 \cap U_2$ has been shown open.
In your version, you just claim without proof that the union of the $B \cap A$ is a union of elements of $\Bbb B$.