$\mathcal{T_B}$ is the intersection of all topologies containing $\mathcal{B}$ Let $\mathcal{B}$ be a basis on a set X, and let $\mathcal{T_B}$ be the topology it generates. Show that
$\mathcal{T_B} =\bigcap \{ \mathcal{T} \subseteq P(X) \mathcal{T}$ is a topology on X and $\mathcal{B} \subseteq \mathcal{T} \}$.That is show that $\mathcal{T_{\mathcal{B}}}$ is the intersection of all topologies that contain $\mathcal{B}$
My attempt at the solution: By definition the Topology $\mathcal{A}$ generated by a basis $\mathcal{B} = \mathcal{A_{\mathcal{B}}} = \{ \bigcup \mathcal{C} : \mathcal{C} \subseteq \mathcal{B} \} \cup \{\emptyset\}$
We want to show that $\mathcal{A_{\mathcal{B}}} = \mathcal{T_B}$ 
First, Since $\mathcal{T_B}$ is the intersection of all topologies containing $\mathcal{B}$ then it is (1) a topology and (2) contains all possible unions of $\mathcal{B}$ since topologies are closed under arbitrary unions. Hence $\mathcal{A_B} \subseteq \mathcal{T_B}$ By definition $\mathcal{T_B}$ is the finest topology that is refined by every topology on X containing $\mathcal{B}$, $\mathcal{A_B}$ is a topology on X containing $\mathcal{B}$ and hence, refines $\mathcal{T_B}$ which implies $\mathcal{T_B} \subseteq \mathcal{A_B}$ So that $\mathcal{A_B} = \mathcal{T_B}$ as required 
 A: The proof is essentially OK. 
Minor quibbles: you don't need the $\{\emptyset\}$, in the definition of $\mathcal{A}_\mathcal{B}$, because we get it for free using $\mathcal{C} = \emptyset$ which has as its union $\emptyset$ as well.
I suppose you already know that $\mathcal{A}_\mathcal{B}$ is a topology, either by the definition of a base or a theorem, and it contains $\mathcal{B}$, because $B = \cup \{B\}$ and $\{B\} \subseteq \mathcal{B}$ for all $B \in \mathcal{B}$. So indeed $\mathcal{A}_\mathcal{B}$ is one of the topologies that we take the intersection of in the definition of $\mathcal{T}_\mathcal{B}$, so $\mathcal{T}_\mathcal{B} \subseteq \mathcal{A}_\mathcal{B}$.  
The other inclusion could be a bit more detailed: let $O \in \mathcal{A}_\mathcal{B}$ so $O = \cup \mathcal{C}$ for some $\mathcal{C} \subseteq \mathcal{B}$. Pick any topology $\mathcal{T}$ that contains $\mathcal{B}$. This $\mathcal{T}$ also contains $\mathcal{C}$ and so also $O = \cup \mathcal{C} \in \mathcal{T}$, as topologies are closed under unions. As $\mathcal{T}$ was arbitrary, $O \in \mathcal{T}_\mathcal{B}$, hence $\mathcal{A}_\mathcal{B} \subseteq \mathcal{T}_\mathcal{B}$. 
