Proposition:
Let $X$ be a locally compact separable metric space, and $\mathscr{B}$ be Borel $\sigma$ -field of $X$.
Then radon measure $m$ on $(X,\mathscr{B})$ is regular.
Here, we define $m$ is radon measure if it is locally finite and inner regular measure.
I can prove if $m$ is finite, but $m$ can be infinite measure.
A separable metric space is second countable, so for Borel set $B$ and a small $\varepsilon$, I tried to construct an open set $U$ s.t.$\ B \subset U$, $m(U-B)<\varepsilon$, but I can't.
So you have a Borel measure $\mu$ that is locally finite (in my definition: every point has an open neighbourhood $O_x$ with $\mu(O_x) < +\infty$) and inner regular: for every Borel set $B$: $$\mu(B)=\sup \{\mu(K): K \text{ compact }, K \subseteq B\}$$
Moreover $X$ is a locally compact separable metric space.
To show is that $\mu$ is outer regular: for all Borel sets $B$:
$$\mu(B)= \inf \{\mu(U): U \text{ open }, B \subseteq U\}$$
which is automatic for all Borel sets $B$ with $\mu(B) = +\infty$ and which for $B$ with finite measure is equivalent to:
$$\forall \varepsilon >0: \exists U \text{ open }: (B \subseteq U) \land \mu(U\setminus B) < \varepsilon$$
As you state in the question, it's not too difficult for finite measures on such spaces. But you can in fact reduce to that finite case (so assume we have proved the finite measure case already as a lemma):
If $B$ is Borel, we can cover it by at most countably many open sets of finite measure (because $X$ is hereditarily Lindelöf (from second countability) and locally finite), say $U_n, n \in \mathbb{N}$. Then define $B_n = B \cap U_n$ and note that, as $U_n$ is also loc. compact sep. metric and $\mu_n = \mu \restriction U_n$ is a finite measure on that space which is still inner regular, $\mu_n$ is outer regular and we can approximate $\mu(B_n)$ by an open subset $V_n$ of $U_n$ (and hence of $X$ too) such that $\mu_n(V_n\setminus B_n) < \frac{\varepsilon}{2^{n+1}}$. Then $V=\bigcup_n V_n$ is as required, by subadditivity of $\mu$ etc.