We have a topological space $X$ with bases $\mathcal{B_1}$ and $\mathcal{B_2}$. For $x\in X$, there is a $B_{1,x}\in\mathcal{B_1}$ and $B_{2,x}\in\mathcal{B_2}$ which contain $x$, by definition of basis.
Since it is a basis, without loss of generality $\mathcal{B_1}$ generates the topology $X$, and so for any open set $U$ in $X$ and $x\in U$ there is a basis element $B_{1,x}$ with $x\in B_{1,x}$ and $B_{1,x}\subseteq U$. Then it follows that such a basis element in $\mathcal{B_2}$, say $B_{2,x}$ contains such $B_{1,x}$, since $B_{2,x}$ is an open set; and likewise for $B_{1,x}$, which is open as well, so that there is a basis element $B'_{2,x}\subseteq B_{1,x}$, etc.
In this fashion, having fixed two bases of $X$, each point of $x$ "induces" a nested collection (or sequence, if it is countable) of 'alternating' basis elements, which may possibly 'terminate' if at some point along this nesting it holds that $B^\alpha_{1,x}=B^\beta_{2,x}$, where $\alpha,\beta$ are in some index sets which index the basis elements of $\mathcal{B_1}$ and $\mathcal{B_2}$, respectively, which participate in this nested collection.
Is this above correct?
Edit This may be a related question: Constructing a local nested base at a point in a first-countable space