Joel has already shown you that none of the standard separation axioms is enough to force a positive answer to your question (by the way, the Sorgenfrey line is even perfectly normal).

A notable case for which your question has a positive answer is that of a topological group.

Define the *$\pi$-weight* of $X$ ($\pi w(X)$) to be the least cardinality of a family $\mathcal{P}$ of non-empty open subsets of $X$ such that for every non-empty open set $U \subset X$ there is $P \in \mathcal{P}$ such that $P \subset U$. It is clear that if $RO(X)$ is a base for $X$ then the density of $RO(X)$ coincides with the $\pi$-weight of $X$. This happens, in particular, if $X$ is a regular space. Now topological groups are regular, and the $\pi$-weight and the weight of a topological group coincide.

(for a proof of that see, for example, *Arhangel’skii, Alexander; Tkachenko, Mikhail*, Topological groups and related structures, Atlantis Studies in Mathematics 1. Hackensack, NJ: World Scientific; Paris: Atlantis Press (ISBN 978-90-78677-06-2/hbk). xiv, 781 p. (2008). ZBL1323.22001.)

Another case in which $\pi w(X)=w(X)$ is when $X$ is a metric space (because the $\pi$-weight is bounded below by the density and bounded above by the weight, and for metric spaces, the weight and the density coincide).