At least under MA+$\neg$CH the answer is negative. It is known that under MA+$\neg$CH the real line contains an uncountable set A such that every subset of A is Borel in A (moreover, it is of type $F_\sigma$ and $G_\delta$).

Now consider the product $X=A\times\mathbb Q$ and let $\mu$ be the (metrizable separable) topology of the space X. Next, consider the topology $\tau$ on $X$, generated by the base $$\mathcal E=\mu\cup\{\{(x,0)\}:x\in A\}.$$

It is easy to see that the $\sigma$-algebra, generated by the base $\mathcal E$ coincides with the Borel $\sigma$-algebra, which coincides with the algebra of all subset of $X$. On the other hand, the open set $A\times\{0\}\in \tau$ cannot be represented as the countable union of basic set from the family $\mathcal E$.