A counterexample to this question (and its locally convex version) is any non-metrizable (locally convex) space $X$, which is hereditarily Lindelof. The hereditary Lindelofness of $X$ implies that any open set is a countable union of basic open sets and this implies that the $\sigma$-algebra generated by any base of the topology coincides with the Borel $\sigma$-algebra.
As for an example of non-metrizable hereditarily Lindelof spaces, take any metrizable separable locally convex space with a weaker non-metrizable topology.
Being a continuous image of a second-countable space, such space will have countable network and hence will be hereditarily Lindelof.
For example, you can take any infinite-dimensional separable Banach space endowed with the weak topology. It will be not metrizable but hereditarily Lindelof.
The function spaces $C_p(X)$ over cosmic spaces $X$ have countable network and hence are hereditarily Lindelof. The function spaces $C_k(X)$ with compact-open topology over $\aleph_0$-spaces $X$ have countable $k$-network and hence are hereditarily Lindelof.
The locally convex space $\mathbb R^\infty$, which is inductive limit of an increasing sequence of finite-dimensional spaces, has countable network and hence is hereditarily Lindelof and not metrizable.
So, there plenty of examples. But one can modify the question replacing the second countability by the hereditary Lindelofness:
Problem. Is a (locally convex) topological space hereditary Lindelof if the $\sigma$-algebra generated by any base of the topology coincides with the $\sigma$-algebra of Borel sets?
Remark. The example given by @Wille Liou in the comment to the original question is hereditarily Lindelof (even hereditarily compact), but not Hausdorff.
In fact, the hereditary Lindelofness admits the following characterization:
Theorem. A topological space $X$ is hereditary Lindelof if and only if for any subspace $Y\subset X$, the $\sigma$-algebra generated by any base of the topology of $Y$ coincides with the Borel $\sigma$-algebra of $Y$.