If $\mathcal B_1\subseteq\tau_2$ is an analytic basis for $\tau_1$, does it follow $\tau_1\subseteq\tau_2$? Let $E$ be a set, $\tau_i$ be a topology on $E$ and $\mathcal B_1\subseteq\tau_2$ be an analytic basis for $\tau_1$.

Are we able to conclude $\tau_1\subseteq\tau_2$?

Intuitively this should be trivially true, but how can we prove it? Please bear with me if this is rather simple, but I'm still trying to get familiar with general topology.

Yes.  This is trivial (he said after leaving the room for half an hour).   $\tau_2$ is closed under unions (because it's a topology), so since everything in $\tau_1$ is a union of elements of $\tau_2$, we have $\tau_1\subseteq\tau_2$.
