*Let X be a metric space. Then every Borel measure μ on X is regular (i.e. for every Borel set B and every ε > 0, there exists a closed set $F_ε$ such that $F_ε ⊂ B$ and μ(B\ $F_ε$) < ε). If X is complete and separable, then the measure μ is Radon .(i.e. for every Borel set B and ε > 0, there exists a compact set $K_ε$ ⊂ B such that μ(B\ $K_ε$) < ε).*
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).