One standard example is the reals numbers times the reals with the discrete topology: $X = \mathbb{R} \times \mathbb{R}_d$.
This is a locally compact metrizable space. The compact subsets intersect only finitely many horizontal lines and each of those non-empty intersections must be compact. A Borel set $E\subset X$ intersects each horizontal slice $E_y$ in a Borel set.
Consider the following Borel measure where $\lambda$ is Lebesgue measure on $\mathbb{R}$:
$$
\mu(E) = \sum_{y} \lambda(E_y).
$$
This is easily checked to define an inner regular Borel measure and its null sets are precisely those Borel sets that intersect each horizontal line in a null set. In particular, the diagonal $\Delta = \{(x,x) : x \in \mathbb{R}\}$ is a null set. However, every open set containing $\Delta$ must intersect each horizontal line in a set of positive measure, so it must have infinite measure and hence $\mu$ is not outer regular.
Now define $\nu$ by the same formula as $\mu$ if $E$ intersects only countably many horizontal lines, and set $\nu(E) = \infty$ if $E$ intersects uncountably many horizontal lines. Now this measure $\nu$ is inner regular on open sets and outer regular on Borel sets.
Finally, you can check that $\mu$ and $\nu$ assign the same integral to compactly supported continuous functions in $X$.
There are some deep relations between:
order structures (e.g. pointwise order of functions),
analytic structures (e.g. the Banach space structure of $C_0(X)$), and
algebraic structures (e.g. pointwise multiplication of functions).
The question under discussion is an example of the relationship between (1) and (2), while (2) and (3) are linked
e.g. via the Theorem asserting that any linear functional $\varphi $ on $C_0(X)$ that preserves multiplication, that is,
$\varphi (fg)=\varphi (f)\varphi (g)$, is necessarily also continuous!
Here is a direct proof of the fact that every positive linear functional on $C_0(X)$ is continuous:
Arguing by contradiction, suppose that for every $n\in {\mathbb N}$, there is $f_n\in C_0(X)$, with $\|f_n\|\leq 1$ and $\varphi (f_n)\geq 4^n$.
Letting $f_n^+$ be the positive part of $f_n$ (that is, $f_n^+(x) = \max\{f_n(x),0\}$), we have that
$$
f_n^+\geq f_n \Rightarrow \varphi (f_n^+)\geq \varphi (f_n)\geq 4^n,
$$
and clearly $\|f_n^+\|\leq 1$.
Setting $f = \sum_{n=0}^\infty 2^{-n}f_n^+$, we then have for all $N\in {\mathbb N}$ that
$$
g_N := \sum_{n=0}^N2^{-n}f_n^+ \leq f,
$$
so $\varphi (g_N) \leq \varphi (f)$. However
$$
\varphi (g_N)= \sum_{n=0}^N2^{-n}\varphi (f_n^+) \geq \sum_{n=0}^N2^{-n}4^n = \sum_{n=0}^N2^n\to \infty ,
$$
a contradiction!
EDIT: Bob's comment and answer below are excellent points and indeed the present answer depends on interpreting $C_0(X)$ as the space of continuous functions vanishing at $\infty$ as opposed to compactly supported. The latter in not complete in the uniform norm so there is no reason for the series representing $f$ to converge. As pointed out, everything is OK with the former, I believe also standard, interpretation of $C_0(X)$.
In case $\varphi$ is instead defined on the space of continuous compactly supported functions, more commonly denoted $C_c(X)$, a similar result may be proved, namely that $\varphi$ is continuous wrt the so called inductive limit topology. This essentialy means that, for each compact subset $K$ of $X$, the restriction of $\varphi$ is continuous on the subspace $C_K(X)$, formed by the continuous functions on $X$, vanishing on the complement of $K$, and equipped with the sup norm.
In the link to the Riesz–Markov theorem provided by the OP there are actually two main results, one for $C_c(X)$ and another for $C_0(X)$. Only the second one involves a necessarily finite measure so the statement by the OP that "$\mu(X)<\infty$" is not compatible with the definition of $C_0(X)$ as the compactly supported functions. I therefore suggest the OP edit the question to resolve this issue.
Best Answer
I imagine you're defining $\mu$ on open sets by $$\mu(U) = \sup \{I(f):\; f \in C_c(X),\; 0 \le f \le 1,\; \text{supp}(f) \subset U\}$$ BTW you left out the $0 \le f \le 1$ in your proof of the finite case.
In the case $\mu(\mathcal O) = \infty$, inner regularity would say that for any positive integer $N$ there is an open set $U$ with compact closure $\overline{U} \subseteq \mathcal O$ and $\mu(U) > N$. Take $f_N \in C_c(X)$ with support contained in $\mathcal O$, $0 \le f \le 1$ and $I(f_N) > N$, and continue as in the finite case.