Completing the converse of the theorem. Definition:-
Let $(X,\mathscr T)$ be a topological space. A basis for $\mathscr T$ is a subcollection $\mathscr B$ of $\mathscr T$ with the property that if $U\in \mathscr T$ then $U=\phi$ or there is a subcollection $\mathscr B'$ of $\mathscr B$ such that $U=\cup \{B:B\in \mathscr B'\}.$
Question:-
A family $\mathscr B$ of subsets of $X$ such that $\mathscr B$ is a basis for the topology $\mathscr T$ on $X$ iff: (1) $X=\{B:B\in \mathscr B\}$,and (2) if $B_1,B_2 \in \mathscr B$ and $x \in B_1 \cap B_2$, then there exists $B\in \mathscr B$ such that $x\in   B$ and $B\subset B_1 \cap B_2$.
My attempt:- 
Assume a family $\mathscr B$ of subsets of $X$ such that $\mathscr B$ is a basis for the topology $\mathscr T$ on $X.$  We know that $X\in \mathscr T.$ We assume that $\mathscr B$ is a basis for the topology $\mathscr T$ on $X.$ there is a subcollection $\mathscr B'$ of $\mathscr B$ such that $X=\cup \{B:B\in \mathscr B'\}\subset \cup \{B:B\in \mathscr B\}\subset X .$ Hence, (1) holds. For proving (2), we have $\mathscr B \subset \mathscr T.$ So, $B_1, B_2\in \mathscr T$. Hence, $B_1 \cap B_2 \in \mathscr T$. So, there exists a subcollection $\mathscr B"$ such that  $B_1 \cap B_2=\{B:B\in \mathscr B"\}.$ So there exists $B\in \mathscr B"\subset \mathscr B$ which contains x and contained in $B_1\cap B_2$.
I have problem on proving on the converse.
Assume (1) and (2). $X\in \mathscr T$, $X=\{B:B\in \mathscr B\}$(given). If we need to prove for $A\neq X$ and $A\neq \phi.$ I am not able to proceed further. How do I complete the proof? Please help me.
 A: Suppose that $\mathcal{B}$ obeys (1) and (2).
Then define $\mathcal{T} = \{ \cup \mathcal{B}': \mathcal{B}' \subseteq \mathcal{B}\}$.
Then $\mathcal{T}$ is a topology on $X$: $X \in \mathcal{T}$ because of (1), while $\mathcal{B}' = \emptyset$ (a valid choice, likewise you don't need the $U= \emptyset$ condition in the question), shows that $\emptyset \in \mathcal{T}$.
That $\mathcal{T}$ is closed under finite intersections follows from (2), intersection of two sets suffice; while closedness under unions is trivial (a union of unions of subfamilies is a union of a subfamily...).
And by definition $\mathcal{B}$ is a base for $\mathcal{T}$.
It's not really a converse in the classical sense: if a family $\mathcal{B}$ of subsets of a set $X$ (no topology yet) obeys the conditions (1) and (2), we can define a topology $\mathcal{T}$ that has $\mathcal{B}$ as a base, which is what I did above. The left to right part, which you did, shows that these conditions are necessary to be a base for a topology, as any base for any topology on $X$ obeys them. (But that topology is not mentioned anymore in the conditions; it's abstracted away as it were.)
A: Regarding the converse: Let $\mathscr B$  satisfy (1) and (2) and  define to be $\{\cup B:B\subset  \mathscr B\}.$ 
(I). Assume that we have shown that $\cup T\in \mathscr T$ whenever $T\subset  \mathscr T.$ 
Regarding the intersection of a finite subset of $\mathscr T$:
(II). From (2),   by induction on $n\in \Bbb N,$ if $\{b_1,...,b_n\} \subset  \mathscr B$ and $p\in \cap_{j=1}^n b_j$ then there exists $b\in \mathscr B$ such that $p\in b\subset \cap_{i=1}^nb_i.$ Therefore $\cap_{i=1}^nb_i \in \mathscr T.$
(III). Now let $n\in \Bbb N$ and $A=\{A_1,...,A_n\}\subset  \mathscr T.$ We want to show that $\cap A\in  \mathscr T.$
For $i=1,...n $ let $A_i=\cup B_i$ where  $B_i\subset \mathscr B.$ We have $$\bullet  \quad\cap A=\cup \{\cap_{i=1}^n b_i: b_1\in B_1\land...\land b_n\in B_n\}.$$
Now if $(b_1\in B_1\land...\land b_n\in B_n)$ then $\{b_i:i=1,...,n\}\subset \mathscr B$ so by (II) we have $\cap_{j=1}^nb_i\in \mathscr T.$ Applying this to $\bullet$ we see that $\cap A$ is the union of a collection of members of $\mathscr T,$ so by (I) we have  $\cap A\in \mathscr T.$
Remarks: (1).In your definition, the phrase " $U=\emptyset$ or " is redundant. A sub-collection $\mathscr B'$ of $\mathscr B$ means a subset  of $\mathscr B.$ This includes the case $\mathscr B'=\emptyset.$ And of course $\cup \emptyset =\emptyset.$..(2). Many authors say base instead of basis. The word basis has (multiple) other meanings in the context of vector spaces, so when discussing topological vector spaces, it may be better to use "base". 
