I think it will help to understand exactly what your after when considering a subbase. Start with choosing a random collection of subsets $S$ of $X$. In general, you know that this collection $S$ does not form a topology. Now comes the fundamental question:
"What do we have to change about $S$ to get a topology?"
Well, for one, arbitrary unions and finite intersections of elements of $S$ ought to again be in $S$. If they are not, we simply add them to $S$ to get a new collection $T$.
Now we also need $X$ and the empty set $\emptyset$ to be in $T$ as well. However, if we add the condition that the union of elements of $S$ is $X$, we know that $X$ will be in $T$ by construction. This is what motivates the definition of a subbase.
In your example, $T=\{X,\emptyset, \{c,d \}, \{b\}\}$ is not a topology since the union
$$\{c,d\}\cup \{b\}=\{c,b,d\}\not \in T.$$
Instead, let's consider
$$T=\{\emptyset, \{b\}, \{c,d \}, \{c,b,d\}, X\}.$$
Does this form a topology? Check the axioms. Is your original example
$$S=\{X,\emptyset, \{c,d \}, \{b\}\}$$
a subbase for T?
In this case, $T$ will consist of all unions of elements of $S$, along with $\emptyset$. Indeed, as you said, the intersection of any two or more elements of $S$ is $\emptyset$, but one element on its own counts as well for an intersection.
Indeed, given any family $\mathcal F$ of sets, the intersection of $\mathcal F$, noted $\bigcap\mathcal F$, is the set of elements that belong to all the elements of $\mathcal F$:
$$\bigcap\mathcal F:=\{x\,\mid\,\forall y\in\mathcal F,\,x\in y\}.$$
A finite intersection of elements of $S$ means $\bigcap\mathcal F$ for a finite $\mathcal F\subset S$. In particular, if $U\in S$, $\bigcap\{U\}=U$.
Best Answer
Other definitions that are equivalent:
Let $X$ be a topological space with topology $\tau$. A subbase of $\tau$ is usually defined as a subcollection $\mathcal{B}$ of $\tau$ satisfying one of the two following equivalent conditions:
[This means that T is the smallest topology containing B: any topology U on X containing B must also contain T.]
[This means that every non-empty proper open set in T can be written as a union of finite intersections of elements of $\mathcal{B}$. Explicitly, given a point x in a proper open set $U$, there are finitely many sets $S_1, \dots, S_n \in \mathcal{B}$, such that the intersection of these sets contains $x$ and is contained in $U$. (wiki)]
A collection of subsets of a topological space that is contained in a basis of the topology and can be completed to a basis when adding all finite intersections of the subsets. (wolfram mathworld)