We can prove it in perturbative string theory but it's probably valid beyond it.

In perturbative string theory, any (continuous) global symmetry has to be associated with a conserved charge which, because of the locality of the physics on the world sheet, implies the existence of a world sheet current $j$ or $\bar j$ or both (left movers vs right movers) whose left/right dimension is $(1,0)$ or $(0,1)$ or both (because its integral has to be a conformally invariant charge). Typically, such a symmetry might be something like the isometry of the target (spacetime) manifold or something on equal footing with it.

It follows that one may also construct operators $j\exp(ik\cdot X) \bar\partial X^\mu$ or $\bar j \exp(ik\cdot X) \partial X^\mu$ or both with a null vector $k$ which have the dimension $(1,1)$, transform as spacetime vectors, and therefore belong to the spectrum of vertex operators of physical states which moreover transform as spacetime vectors, i.e. they're the gauge bosons (the $X$ only go over the large spacetime coordinates). One would need to prove that the multiplication of the operators doesn't spoil their tensor character but it usually holds.

Consequently, any would-be global symmetry may automatically be shown to be a gauge symmetry as well.

The argument above only holds for the gauge symmetries that transform things nontrivially in the bulk of the world sheet. But even symmetries acting on the boundary degrees of freedom, i.e. the Chan-Paton factors, obey the same requirement because one may also construct (open string) vertex operators for the corresponding group that transform properly.

We don't have a universal non-perturbative definition of string theory but it's likely that the conclusion holds non-perturbatively, too.

In some moral sense, it holds for discrete symmetries as well even though discrete symmetries don't allow gauge bosons.

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