If $\mathcal{B}$ is a base of a topology space $\left(X,\tau\right)$, then the Borel $\sigma$-algebra is generated by $\mathcal{B}$?

Let $\left(X,\tau\right)$ a topology space and $\mathcal{B}$ a base of the topology, my question is:

The Borel $\sigma$-algebra is generated by $\mathcal{B}$ ?

Not necessarily. For example, consider the discrete topology on $\Bbb R,$ which has $$\mathcal B=\bigl\{\{x\}:x\in\Bbb R\bigr\}$$ as a base. The $\sigma$-algebra generated by $\mathcal B$ is the set of all subsets $A$ of $\Bbb R$ such that either (1) $A$ is finite or countably infinite, or (2) $\Bbb R\setminus A$ is finite or countably infinite. However, the appropriate Borel $\sigma$-algebra is the power set of $\Bbb R,$ so for example, $[0,1]$ is an element of the Borel $\sigma$-algebra that is not an element of the $\sigma$-algebra generated by $\mathcal B$.

To some extent, it depends on the topology in question--for example, given any second-countable topology and any base for that topology, the $\sigma$-algebra generated by the base is the Borel $\sigma$-algebra (and I suspect that only second-countable topologies necessarily have this property, unless we assume the Axiom of Choice). On the other hand, it also depends on what base you choose, since every topology has itself as a base.

As Cameron Buie pointed out, the claim is not true in general.

However, it is true if $\tau$ is second-countable and $\mathcal B$ is a countable topological basis (in particular, it is true in separable metric spaces). To see this, suppose that $U\in\mathcal \tau$ is any open set. Then, there exists a subcollection $\mathcal B_{U}\subseteq\mathcal B$, necessarily countable, such that $$U=\bigcup_{B\in\mathcal B_U}B.$$ This union is countable, so that $U\in\sigma(\mathcal B)$. Since $U$ is an arbitrary open set, it follows that $\tau\subseteq\sigma(\mathcal B)$, and, in turn, $\sigma(\tau)\subseteq\sigma(\mathcal B)$.

The other direction $\sigma(\mathcal B)\subseteq\sigma(\tau)$ is easy, given that $\mathcal B\subseteq\tau$. It follows that the Borel $\sigma$-algebra $\sigma(\tau)$ equals the $\sigma$-algebra generated by the topological basis $\mathcal B$.

In fact, if $\tau$ is second-countable, then any topological basis $\mathcal C$, countable or not, will generate the Borel $\sigma$-algebra. This is because then $\mathcal C$ contains a countable subcollection $\mathcal B\subseteq\mathcal C$ that is still a basis (see here for details). As shown above, $\sigma(\tau)=\sigma(\mathcal B)\subseteq\sigma(\mathcal C)$, and the other direction $\sigma(\mathcal C)\subseteq\sigma(\tau)$ is again easy.