Can a similar result to this question be true for a topological space that has a pseudo basis "A collection of non-empty open sets (in a topological space X) such that every non-empty open set of X contains one of these is called a pseudo-basis"
Edit: Question: If $\mathcal B$ is a countable pseudo base for a topological space $(X,\tau)$, then $\sigma(\mathcal B)=\sigma(\tau)$, where $\sigma(*)$ is the sigma algebra generated by $(∗)$.
Suppose $X$ is a second-countable Hausdorff space and that $x,y$ are two distinct limit points of $X$. Let $\{U_n\}$ be any countable basis for $X$ and let $V_n = U_n \setminus \{x,y\}$, so that $V_n$ is open, nonempty, and $V_n \subset U_n$. Then $\{V_n\}$ is a countable pseudo-basis. However, every set in the $\sigma$-algebra generated by $\{V_n\}$ either contains $\{x,y\}$ or is disjoint from it. In particular, this $\sigma$-algebra does not contain the Borel set $\{x\}$.