Consider the following definitions:
Let $(X, \tau)$ be a topological space and let $(X, \Sigma, \mu)$ be a measure space such that $\tau \subseteq \Sigma$.
We say that $\mu$ is a Radon Measure on $X$ if and only if it is locally finite (i.e. every point in $X$ has an open neighborhood with finite measure) and inner regular for open sets.
We say that $\mu$ is a Riesz Measure on $X$ if and only if $(X, \tau)$ is locally compact, Hausdorff and $\mu$ is finite on compact subspaces, outer regular and inner regular for open and finite measure sets.
The definition of a Radon Measure is taken from Wikipedia (https://en.wikipedia.org/wiki/Radon_measure#Definitions) and the definition of a Riesz Measure is drawn from the properties of the measure obtained by applying the Riesz Representation Theorem (see for instance Rudin "Real and Complex Analysis", Theorem 2.14).
I would like to know what are minimal conditions to impose on a Radon Measure on a locally compact, Hausdorff space in order to make it into a Riesz Measure.
Indeed I have read somewhere, without proof, that a Radon Measure on a locally compact, Hausdorff space is automatically outer regular but I have only succeeded in proving that it is outer regular for compact subspaces.
Any comment or answer is much appreciated and let me know if I can explain myself clearer.
The claim that you found is false: on some locally compact Hausdorff spaces, it is possible to construct a locally finite measure $\mu$ which is inner but not outer regular. Indeed, consider the space $$(X, \tau)=\mathbb{R} \times (\mathbb{R}, \tau_d)$$ where $\tau_d$ denotes the discrete topology (that is, every subset of $\mathbb{R}$ is open), and define the measure $$\mu(E):=\sum_{y \in \mathbb{R}} \mathcal{L}^1(E_y)$$ where $\mathcal{L}^1$ denotes the 1-dimensional Lebesgue measure and $E_y:=\{x \in \mathbb{R}: (x,y) \in E\}$ are the horizontal sections of $E$.
Then one can easily check that $\mu$ is locally finite. It is also inner regular: for example, if $\mu(E)<\infty$ then $\mathcal{L}^1(E_y) >0$ only for countably many $y$, thus the series defining $\mu(E)$ can be approximated by a finite sum of the form $$\sum_{i=1}^N \mathcal{L}^1(E_{y_i}).$$ Finally, approximate each $E_{y_i}$ with a compact set $K_i$ from the inside using the inner regularity of $\mathcal{L}^1$, and take $\displaystyle K:=\bigcup_{i=1}^N K_i \times \{y_i\}$. The case $\mu(E)=\infty$ is similar.
However, $\mu$ is not outer regular. Indeed, the line $L:=\{(x,y): y=x\}$ is a $\mu$-null set, but every open set $U$ containing $L$ must intersect each horizontal line in a set with positive 1-dimensional Lebesgue measure (as the horizontal lines are open sets too for this topology), hence $\mu(U)=\infty$.
Now, coming to your main question: I think that the most reasonable condition to add in order to ensure the outer regularity of $\mu$ is the $\sigma$-compactness of $(X, \tau)$. Indeed, as you say this property is true for $X$ compact and thus also is $X$ can be written as countable union of compacts. The $\sigma$-compactness follows authomatically, for example, if the topology $\tau$ is second-countable.