A basis may not be a topology itself? Is my understanding correct here -
Let $X = \{a, b, c\}$
Let $\beta = \{\{a\}, \{a, b\}, \{a, c\}\}$
Now $\beta$ is a basis as $X = \{a, b\} \bigcup\{a, c\}$, and for any $B_1, B_2 \in \beta$ we have that $B_1 \bigcap B_2 \in \beta$.
Hence $\beta$ will generate a topology on $X$ but $\beta$ itself is not a topology on $X$ as a topology is closed under arbitrary unions and $\{a, b\} \bigcup\{a, c\} \notin \beta$.
So is my understanding correct?
 A: Yes, your understanding is correct.
A: Every topology needs to have X and the empty set. Your $\beta$ does not have these, so it is not a topology. This generates a new question: If we add X and the empty set to the basis, does then we end up with a topology?
A: To expand upon the above answers a little more, your understanding is exactly correct, and here are the reasons for doing what we do with bases and subbases.
For a basis, we know that it has the property that it covers the space, and that for every pair of basis element $B_1$ and $B_2$, given $x$ in their intersection $B_1\cap B_2$, there is a basis element $B_3$ such that $x\in B_3\subset B_1\cap B_2$.  This isn't exactly the necessity that a topology be closed under finite-intersection, but it is very close.  The main issue is that, as you noticed, it need not be closed under arbitrary union.  Thus, the topology generated by a basis is precisely the collection of sets formed by these arbitrary unions (and this corrects the finite-intersection issue as well).
For a subbasis, we only know that it covers the space, so we need to both introduce the finite intersections AND the arbitrary unions.
Finally, if we have a subbasis or basis that already form a topology, then the topology generated by them will be the same topology.
