Let $\mathcal S, \mathcal S'$ be two subbases for the topologies $\tau , \tau'$, respectively. What would be a mathematical statement $(A)$ in terms of subbases $\mathcal S,\mathcal S'$ which is equivalent to $\tau\subset \tau'$ ?
There is an answer here When do two subbases generate the same topology, but the given condition is not equivalent to $\tau\subset \tau'$; because $\tau\subset \tau'$ implies the given condition, but I don't find how the converse is true.
The quoted answer is actually necessary and sufficient:
$\tau \subseteq \tau'$ iff $$\forall S \in \mathcal{S}: \forall x \in S: \exists S_1, \ldots S_n \in \mathcal{S}': x \in \bigcap_{i=1}^n S_i \subseteq S\tag{1}$$
If $\tau \subseteq \tau'$ then $(1)$ hold, as $S$ is open in $\tau$ and so open in $\tau'$ so for every $x \in S$ such a basic set (from the base generated from $\mathcal{S}'$) must exist. So the condition is necessary.
If, OTOH, $(1)$ holds, it's clear by the same reasoning that every $S \in \mathcal{S}$ is open in $\tau'$ which implies that $\tau$ (the minimal topology that contains $\mathcal{S}$) is contained in $\tau'$ (which is one of these topologies).
I think this is the most natural condition to check this inclusion.
