What does this property of the basis try to generalize? Given a topological space $X$ and $x  \subset B_1 \cap B_2$ where $B_1$ and $B_2$ are basis elements then there exists a basis $B_3$ such that $ x \in B_3 \subset B_1 \cap B_2$.Is it trying to generalize the connectedness property of a topological space ?
 A: No, it's not related to connectedness. Given a set $X$, a basis $\mathcal B$ is a collection of subsets of $B$ such that

*

*$X\in B$ or $\mathcal B$ covers $X$ (the important thing is that $X$ can be written as a union of elements from $\mathcal B$)


*For every $B_1,B_2\in\mathcal B$ and $x\in B_1\cap B_2$, there is a $B_3\in \mathcal B$ such that $x\in B_3\subseteq B_{1}\cap B_2$
The second condition implies that $B_1\cap B_2$ is equal to $$\cup\{B\in\mathcal B|B\subseteq B_1\cap B_2\}$$
Now, if you have a basis $\mathcal B$, you can define the set $$\tau=\{\cup \mathcal A\text{ for all }\mathcal A\subseteq\mathcal B\}$$(where $\cup\mathcal A=\bigcup_{A\in\mathcal A}A$) and prove that $\tau$ is a topology:

*

*Since $\cup\emptyset=\emptyset$, then $\emptyset\in\tau$, while the first condition on the definition above implies that $X\in\tau$


*Clearly $\tau$ is closed under arbitrary unions


*Finally, you use the second condition in the definition above to prove that $\tau$ is closed under finite intersection: Take $\bigcup_iB_i,\bigcup_jA_j\in\tau$ $$\left(\bigcup_iB_i\right)\cap\left(\bigcup_jA_j\right)=\bigcup_{i,j}B_i\cap A_j=\bigcup_{i,j}\bigcup\{C\in\mathcal B|C\subseteq B_i\cap A_j\}$$
So that condition ensures that you can use the basis to form a topology in a "simple" way. If you drop such a condition, you get a subbase, which you can still use to construct a topology by first forming a base, then using that base to form a topology.
Edit To answer the first comment: The notion of subbase is obtained from the notion of base by dropping the second condition. So being a subbase is a weaker notion of being a base, that is, a base is always a subbase, but not the contrary. But, if you have a subbase $\mathcal S$, the set $$\{\cap A|A\subseteq\mathcal S,A\text{ is finite}\}$$(that is, the set of finite intersections of elements in $\mathcal S$) is a base. For example, consider a countable product $$X=\prod_{i\in\mathbb N}X_i$$ and let $\pi_i$ be the projection $X\rightarrow X_i$, then a subbase for the product topology on $X$ is $$\{\pi^{-1}_i(O)|i\in\mathbb N,O\text{ open in }X_i\}$$ this subbase induces the base $$\{U\subseteq X|\pi_i(U)\neq X_i\text{, for finitely many }i\}$$
If you have a map $f:(X,\tau)\rightarrow(Y,\mu)$ between two topological spaces and a base $\mathcal B$ (resp. a subbase $\mathcal S$), then $f$ is continuous if and only if $f^{-1}(U)$ is open, for all $U\in\mathcal B$ (resp. for all $U\in\mathcal S$). Many topologies are pretty difficult to describe directly (that is, describe a general open set in that topology), so usually we simply give a base or subbase for the topology we are considering (like the product topology above, which is described using the above basis), since having the base (or a subbase) for a topology on $Y$ is enough to check if a function $X\rightarrow Y$ is continuous.
A: If you think of topological spaces as a generalization of metric spaces, perhaps the following is enlightening: in a metric space $(X, d)$, the basic open sets are the balls
$$
B(a, r) = \{x \in X \mid d(a, x) < r\}.
$$
Then the criterion that you are asking about says: suppose we have two open balls $B(a_1, r_1)$ and $B(a_2, r_2)$, such that $B(a_1, r_1) \cap B(a_2, r_2)$ contains an element $x$, then there is another ball contained in the intersection that also contains $x$.
To see this, write $b_1 = d(a_1, x)$ and $b_2 = d(a_2, x)$. Then, we can choose the ball
$$
B(x, \min(r_1 - b_1, r_2 - b_2)).
$$
Note that $\min(r_1 - b_1, r_2 - b_2) > 0$ because $x \in B(a_1, r_1)$ and $x \in B(a_2, r_2)$. Now let us check that
$$
B(x, \min(r_1 - b_1, r_2 - b_2)) \subseteq B(a_1, r_1) \cap B(a_2, r_2).
$$
Take $y$ in the left hand set. Then
$$
d(y, a_1) \leq d(y, x) + d(x, a_1) < r_1 - b_1 + b_1 = r_1,
$$
and the same works when replacing all 1s in that line by 2s.
Thus: you could see this as a generalization of the triangle inequality.
