The conditions stated in the question seem mouthful and a bit arbitrary, so let me provide some backgrounds.
Definition
Let $\mu$ be a Borel measure on a topological space. We say:
$\mu$ is outer regular on a Borel set $E$ if $\mu(E) = \inf\{\mu(U) : E\subseteq U \text{ is open}\}$,
$\mu$ is inner regular on a Borel set $E$ if $\mu(E) = \sup\{\mu(K) : E\supseteq K \text{ is compact}\}$ (Some authors call this tight),
$\mu$ is a Radon measure if $\mu$ is finite on all compact sets, outer regular on all Borel sets and inner regular on all open sets,
$\mu$ is a regular measure if $\mu$ is finite on all compact sets and both outer regular and inner regular on all Borel sets.
The subtle difference between a Radon measure and a regular measure is annoying. Fortunately, every $\sigma$-finite Radon measure on a locally compact Hausdorff space is automatically regular:
Theorem 1
Let $X$ be a locally compact Hausdorff space. Then every Radon measure on $X$ is inner regular on all of its $\sigma$-finite sets. In particular, every $\sigma$-finite Radon measure on $X$ is regular.
Corollary 1
Let $X$ be a locally compact Hausdorff space. A finite Borel measure on $X$ is regular if and only if it is outer regular on all Borel sets and inner regular on all open sets.
Next, by using Riesz representation theorem on locally compact Hausdorff spaces, one can prove the following:
Theorem 2
Let $X$ be a locally compact Hausdorff space. If every open set in $X$ is $\sigma$-compact, then every Borel measure on $X$ that is finite on compact sets is regular.
A special case is a separable metric space:
Corollary 2
Let $X$ be a locally compact, separable metric space. Then every finite Borel measure on $X$ is regular.
The proof of this corollary relies on the following general topological result:
Lemma
A locally compact metric space is $\sigma$-compact if and only if it is separable, in which case every open set is $\sigma$-compact.
In case local compactness is not given, one still has the following result:
Theorem 3
Let $X$ be a separable complete metric space. Then every finite Borel measure on $X$ is regular.
Note that a locally compact metric space can be given a compatible complete metric, so Theorem 3 also implies Corollary 2.
Another special case of Theorem 2 is a second-countable space:
Corollary 3
Let $X$ be a second-countable and locally compact Hausdorff space. Then every finite Borel measure on $X$ is regular.
Proof: A locally compact Hausdorff space is topologically regular, so by Urysohn's metrization theorem, $X$ is metrizable. On the other hand, a second-countable space is separable. Thus, $X$ is locally compact, separable and metrizable, so Corollary 2 applies.
Finally, in view of the above results, my question is:
Question
Let $X$ be a locally compact Hausdorff, $\sigma$-compact and separable space, which may or may not be metrizable. Is every finite Borel measure on $X$ regular? If not, please give a counter-example.
Best Answer
No. Counterexamples include $X = \{0,1\}^\kappa$ or $[0,1]^\kappa$ for any $\aleph_0 < \kappa \le \mathfrak{c}$. These spaces are compact Hausdorff (Tikhonov's theorem) and are separable (Hewitt-Marczewski-Pondiczery).
(This special case of H-M-P can also be proved directly. For example, with $X =[0,1]^\kappa$, identify $\kappa$ with a subset of $[0,1]$, so that $X$ is a space of functions from $[0,1]$ to itself; then it is easy to show that the set of polynomials with rational coefficients is dense in $X$.)
Now the uncountable successor ordinal $\omega_1 + 1$, with its order topology, can be embedded in $X = \{0,1\}^\kappa$ or $X = [0,1]^\kappa$. See Fremlin, Measure Theory, 434K(d) (thanks to Robert Furber for the reference). One can also follow a construction similar to the Stone-Čech compactification and embed $\omega_1 + 1$ into $[0,1]^{C(\omega_1+1, [0,1])}$, noting that $|C(\omega_1 + 1, [0,1])| = \mathfrak{c}$. Either way, call the embedding $\phi$.
Let $\nu$ be the famous Dieudonné measure on $\omega_1 + 1$, where $\nu(B) = 1$ if $B$ contains a closed unbounded subset of $\omega_1$ and $\nu(B) = 0$ otherwise. (See Fremlin 411Q and 4A3J for details.) Then the pushforward $\mu = \nu \circ \phi^{-1}$ is a finite Borel measure on $X$ which is not regular.
Specifically, since $\phi$ is an embedding and $\omega_1$ is open in $\omega_1 + 1$, then the image $A = \phi(\omega_1)$ is a relatively open subset of the compact set $\phi(\omega_1 + 1)$, so that $A$ is the intersection of an open set and a closed set in $X$, and in particular is Borel. Moreover $\mu(A) = \nu(\omega_1) = 1$. But if $K$ is any compact subset of $A$, then $\phi^{-1}(K)$ is a compact subset of $\omega_1$, hence $\mu(K) = \nu(\phi^{-1}(K)) = 0$. Thus inner regularity fails.
This example is also Exercise 7.14.130 of Bogachev's Measure Theory.