Let $(X, \beta, \mu)$ be a measure space, and $(X, \tau)$ be a Hausdorff topological space such that:

1. $\mathcal{B}(\tau)\subset\beta$; where $\mathcal{B} (\tau)$ is the Borel set generated by $\tau$.
2. There is a sequence $$ (K_n)_{n\in \mathbb{N}} \in X $$ of compact sets such that $X$ is equal to the union of all those sets in the sequence.
3. The measure $\mu(K)$ is finite, for all $K$ compact that is in $X$.

I want to prove that, if $\mu$ is regular, then for each $A\in\beta$, and for all $\epsilon>0$, there exists an open set $U \supset A$ such that: $$\mu (U-A) \leq \epsilon$$.

To solve the problem, do we need full regularity or just outer regularity? And why does the space have to be specifically Hausdorff? Can the problem still be solved with a stronger/weaker separation axiom?
In general, what restrictions given in the problem statement can we remove/add while still having a similar solution?

Let $(X, \beta, \mu)$ be a measure space, and $(X, \tau)$ be a Hausdorff topological space such that:

1. $\mathcal{B}(\tau)\subset\beta$; where $\mathcal{B} (\tau)$ is the Borel set generated by $\tau$.
2. There is a sequence $$ (K_n)_{n\in \mathbb{N}} \in X $$ of compact sets such that $X$ is equal to the union of all those sets in the sequence.
3. The measure $\mu(K)$ is finite, for all $K$ compact that is in $X$.

I want to prove that, if $\mu$ is regular, then for each $A\in\beta$, and for all $\epsilon>0$, there exists an open set $U \supset A$ such that: $$\mu (U-A) \leq \epsilon$$.

To solve the problem, do we need full regularity or just outer regularity? And why does the space have to be specifically Hausdorff? Can the problem still be solved with a stronger/weaker separation axiom?
In general, what restrictions given in the problem statement can we remove/add while still having a similar solution?

Edit: The following is a relevant post about regular measures.
Is every finite Borel measure on a locally compact Hausdorff, $\sigma$-compact and separable space automatically regular?