The definition of subbasis given to me is the following:
Given a topological space $(X,T)$ a subset $S\subset T$ is a sub-basis for $T$ if the set $B=\{\bigcap F:\text{$F$ is finite, and $F\subset S$}\}$ forms a basis for $T$.
Then the statement
For a given collection $S$ of subsets of $X$, there is always a topology $T_S$ for which $S$ is a sub-basis for $T_S$
But I'm not convinced that this is true. Consider for example $X=\{0,1\}$ and $S=\{\{0\}\}$. Then $B=\{\emptyset,\{0\}\}$, but I know that for a set $B\subset \mathcal{P}(X)$ to be a basis for some topology on $X$, it must satisfy the two conditions
$\bigcup B=X$ and for every $B_1,B_2\in B$ there exists some $B'\subset B$ such that $B_1\cap B_2=\bigcup B'$
The example I have does not satisfy $\bigcup B =X$. So it cannot possibly be a basis for some topology.
And even in abstract setting, I fail to see how any general collection $S$ can generate $B$ that satisfies above two conditions: I'm sure I'm somehow confused about the statements or concepts - it'd be nice if someone can point it out
If $S$ is a collection of subsets of $X$, then among the finite intersections from $S$ is the empty intersection, which is equal to $X$.
In your example, $B = \{\{0\},X\}$, not $\{\{0\},\emptyset\}$.