Assume we have two topological spaces $(X, \mathcal{O}_1)$ and $(X, \mathcal{O}_2)$ with two subbases $\mathcal{S}_1$ and $\mathcal{S}_2$, resp.
Now I asked myself whether the following statement holds:
If $\mathcal{S}_1 \subseteq \mathcal{O}_2$ and $\mathcal{S}_2 \subseteq \mathcal{O}_1$, then we have $\mathcal{O}_1 = \mathcal{O}_2$
I think I proved it, but since I'm new to topology I'm not quite sure whether my argument is correct. My idea:
Since $\mathcal{S}_1$ is a subbase of $\mathcal{O}_1$, $\mathcal{S}_1 \cup \mathcal{S}_2 \supseteq \mathcal{S}_1$ is also a subbase of $\mathcal{O}_1$.
By the same argument $\mathcal{S}_1 \cup \mathcal{S}_2 \supseteq \mathcal{S}_2$ is a subbase of $\mathcal{O}_2$.
So $\mathcal{O}_1$ and $\mathcal{O}_2$ have a common subbase, therefore they must be equal.
Edit: also, can we show the same thing if we have two spaces $X$ and $Y$ and a priori don't know if $X=Y$? Because I think this follows from $X\in \mathcal{O}_1\cap \mathcal{O}_2$ and $Y \in \mathcal{O}_1\cap \mathcal{O}_2$.
$\mathcal{S}_1$ is a subbase for $\mathcal{O}_1$, which means that $\mathcal{O}_1$ is the smallest topology on $X$ that contains $\mathcal{S}_1$ as a subset.
If now $\mathcal{S}_1 \subseteq \mathcal{O}_2$, that minimality of $\mathcal{O}_1$ (as $\mathcal{O}_2$ is a topology that contains $\mathcal{S}_1$) gives us that $$\mathcal{O}_1 \subseteq \mathcal{O}_2$$ and the reverse inclusion follows in the same way, mutatis mutandis.
So we indeed have equality of topologies.