Inner regularity property of Radon measures in metric spaces Let us agree to say that $\mu$ is a Radon measure on a metric space $X$ if it is a Borel measure which is finite on compact subsets and is such that:


*

*Every measurable subset $A$ is outer regular, meaning that $$\mu(A)=\inf\{\mu(V)\ |\ A\subset V,\ V\ \text{is open}\};$$

*Every open subset $U$ is inner regular, meaning that $$\mu(U)=\sup\{\mu(K)\ |\ K\subset U,\ K\ \text{is compact}\}.$$


The present question regards inner regularity. Indeed, as I read in Evans-Gariepy's Measure theory and fine properties of functions, Theorem 4, Chapter 1 (*), if $X=\mathbb{R}^n$ then every measurable subset is automatically inner regular. Is this a property of $\mathbb{R}^n$ alone? Formally:

Question. Which metric spaces have the property that for any Radon measure every measurable subset is inner regular? 


(*) Notation and conventions in this book are a bit different from the ones of the present post. 
EDIT. Specifically, in Evans-Gariepy's book a measure is a extended-real valued set function which is monotone and subadditive (usually, this is called a outer measure). A measurable set is one which satisfies Caratheodory's criterion:
$$E\ \text{is measurable} \iff \forall T\subset X,\ \mu(T)=\mu(T\cap E)+
\mu(T\cap E^c).$$
A Radon measure is a (outer) measure which is:


*

*Borel regular, meaning that every Borel set is measurable and every set (even nonmeasurable ones) is contained in a Borel set of the same (outer) measure;

*Finite on compact subsets.


The aforementioned Theorem 4 of Chapter 1 says that, given a Radon (outer) measure on $\mathbb{R}^n$, every set (measurable or not) is outer regular and every measurable set is inner regular. 
Remark 1. If $\mu$ is a Borel regular measure on $X$ such that every Borel set is inner regular, then every measurable subset of finite measure is inner regular. Indeed, if $M\subset X$ is measurable and has finite measure, then by applying two times the Borel regularity property we can get a Borel set $M'$ which is contained in $M$ and has the same measure as $M$. 
Remark 2. In particular, if $X$ is locally compact and is expanding union of compact sets, as in Micheal's kind answer below, and if $\mu$ is a Borel regular Radon (outer) measure, I believe that every (Caratheodory) measurable subset is inner regular. Indeed let $M\subset X$ be measurable. If $\mu(M)<\infty$ we are done. If $\mu(M)=\infty$ then we can write it as an expanding union of sets of finite measure: $M=\cup_1^\infty M_j$. Every $M_j$ contains a compact $K_j$ such that $\mu(K_j)\ge \mu(M_j)-1$. Letting $j\to \infty$, $\mu(M_j)\to \mu(M)=\infty$ and so $\mu(K_j)\to \infty$ too. This proves the claim.
 A: Every finite measure on a topological space is outer regular if and only if it is inner regular, so compactness would be sufficient (and every finite measure on metric space is automatically both inner and outer regular). Another sufficient condition is that the space is the union of an increasing sequence of compact sets, which covers the case of $\mathbb{R}^n$.
One can look this up in: Aliprantis & Border 2006, "Infinite Dimensional Analysis" in section 12.1.
Edit: Lurker has pointed out that aliprantis and Border use the weaker notion of inner regularity of apprximating sets from below by closed sets. What is called inner regularity here, they call tightness. It is automatic for finite measures on Polish spaces but not all metric spaces.
A: In every locally compact $\sigma$-compact space a Radon measure, as you defined it, is inner regular (on all - not just open - measurable sets). This is Corollary 7.6 from Real Analysis by Folland (who uses the same definitions as you). You don't need it to be a metric space.
(Although it's nice to note that if it is a metric space, in addition to being locally compact $\sigma$-compact, then all locally finite Borel measures satisfy these properties by Corollary 7.8 of the same book).
Note: this might be exactly what Greinecker said, but it seems he referred to a book using a different definition, so I thought this might be helpful.
