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Unfortunately, I can only provide a reference but no ideas since I don't have the paper. In "On the problem of generating sigma-algebras by topologies", Statist. Decisions 2 (1984), 377-388, Albert Ascherl shows (at least according to the summary to be found on MathSciNet) that there are $\sigma$-algebras which can't be generated by a topology.

Robert Lang (same journal 4 (1986), 97-98) claims to give a shorter proof.

As suggested by Joel, I add the ideas of Lang's example. The underlying space is $\Omega= 2^{\mathbb R}$, that is the space of all indicator functions, and the $\sigma$-algebra is $\mathcal A = \bigotimes_{\mathbb R} \mathcal P$ where $\mathcal P$ is the power set of the two element set. It is generated by the system $\mathcal E$ of the "basic open sets" of the product topology (prescribed values in a finite number of points). This generator has the cardinality $c$ of the continuum and since the generated $\sigma$-algebra can be obtained in $\omega_1$ (transfinite) induction steps the cardinality of $\mathcal A$ is also $c$. On the other hand, if $\mathcal T$ is a topology with $\mathcal A=\sigma(\mathcal T)$ then $\mathcal T$ separates points (this should follow from the "good sets principle"), in particular, for two distinct points of $\Omega$ the closures of the corresponding singletons are distinct. Hence $\mathcal T$ has at least $|\Omega|=2^c$ elements.

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Unfortunately, I can only provide a reference but no ideas since I don't have the paper. In "On the problem of generating sigma-algebras by topologies", Statist. Decisions 2 (1984), 377-388, Albert Ascherl shows (at least according to the summary to be found on MathSciNet) that there are $\sigma$-algebras which can't be generated by a topology.

Robert Lang (same journal 4 (1986), 97-98) claims to give a shorter proof.