*Let X be a metric space. Then every Borel measure μ on X is regular. If X is complete and separable, then the measure μ is Radon.*

This result is proved on p. 70 in  *"Measure Theory" vol. 2, Springer-Verlag, Berlin
2007,  by  V. I. Bogachev* (Theorem 7.1.7.)  An example of a regular Borel measure which is not tight is provided on the same page (Example 7.1.6).

P.S. Just a comment on the answer by Ian Morris: *tightness of a regular Borel measure on X may fail even if X is a separable metric space*. For example, we may take a restriction of the standard Lebesgue measure to a nonmeasurable subset of the interval $[0, 1]$ with zero inner measure and unit outer measure (endowed with the usual metric).