I think the following is a counterexample.

Consider the Stone–Čech compactification $\beta \mathbb{N}$. Fix some $x \in \beta \mathbb{N} \setminus \mathbb{N}$ and set $S = \beta \mathbb{N} \setminus \{x\}$, which is locally compact Hausdorff and not compact. Let $\mu$ be the Borel measure on $S$ which puts mass $2^{-n}$ on each point $n \in \mathbb{N} \subset S$. This measure is regular, and its support is all of $S$ (which is non-compact), since $\mathbb{N}$ is dense in $S$.

Suppose $\mathcal{H}$ is an infinite disjoint family of Borel sets having positive measure. Then each $H \in \mathcal{H}$ must contain at least one point of $\mathbb{N}$, so write $\mathcal{H} = \{H_1, H_2, \dots\}$, and for each $k$ choose some $n_k \in H_k \cap \mathbb{N}$. Set $A = \{n_1, n_3, n_5, \dots\}$ and $B = \{n_2, n_4, n_6, \dots\}$. I claim that either $A$ or $B$ is contained in a compact set $K$.

To see this, take some function $f : \mathbb{N} \to \{0,1\}$ which is $0$ on $A$ and $1$ on $B$, and extend it to a continuous $\hat{f} : \beta \mathbb{N} \to \{0,1\}$. Suppose that $\hat{f}(x) = 1$. Then the set $K = \hat{f}^{-1}(\{0\})$ is closed in $\beta \mathbb{N}$, hence compact, and contains $A$ but not $x$. So $K$ is also compact in $S$, and since it contains $A$, it intersects $H_1, H_3, H_5, \dots$ with positive measure. If instead we had $\hat{f}(x) = 0$, then just interchange $A$ and $B$, and get a compact set $K$ intersecting $H_2, H_4, H_6,\dots$.

I think the following is a counterexample.

Consider the Stone–Čech compactification $\beta \mathbb{N}$. Fix some $x \in \beta \mathbb{N} \setminus \mathbb{N}$ and set $S = \beta \mathbb{N} \setminus \{x\}$, which is locally compact Hausdorff and not compact. Let $\mu$ be the Borel measure on $S$ which puts mass $2^{-n}$ on each point $n \in \mathbb{N} \subset S$. This measure is regular, and its support is all of $S$ (which is non-compact), since $\mathbb{N}$ is dense in $S$.

Suppose $\mathcal{H}$ is an infinite disjoint family of Borel sets having positive measure. Then each $H \in \mathcal{H}$ must contain at least one point of $\mathbb{N}$, so write $\mathcal{H} = \{H_1, H_2, \dots\}$, and for each $k$ choose some $n_k \in H_k \cap \mathbb{N}$. Set $A = \{n_1, n_3, n_5, \dots\}$ and $B = \{n_2, n_4, n_6, \dots\}$. I claim that either $A$ or $B$ is contained in a compact set $K$.

To see this, take some function $f : \mathbb{N} \to \{0,1\}$ which is $0$ on $A$ and $1$ on $B$, and extend it to a continuous $\hat{f} : \beta \mathbb{N} \to \{0,1\}$. Suppose that $\hat{f}(x) = 1$. Then the set $K = \hat{f}^{-1}(\{0\})$ is closed in $\beta \mathbb{N}$, hence compact, and contains $A$ but not $x$. So $K$ is also compact in $S$, and since it contains $A$, it intersects $H_1, H_3, H_5, \dots$ with positive measure. If instead we had $\hat{f}(x) = 0$, then just interchange $A$ and $B$, and get a compact set $K$ intersecting $H_2, H_4, H_6,\dots$.

It is true if $S$ is metrizable. Fix a compatible metric $d$. Since $M$ is not compact, there is a sequence $x_n \in M$ with no convergent subsequence; since $M$ is closed, no subsequence converges in $S$ either. Let $B_n$ be disjoint open balls centered at $x_n$, having radius at most $1/n$. Since each $x_n$ is in the support, each $B_n$ has positive measure, so take $\mathcal{H} = \{B_n\}$. Now if $K$ meets infinitely many $B_n$ (at all), then there is a sequence $y_{n_k} \in K \cap B_{n_k}$. If this has a convergent subsequence $y_{n_{k_j}} \to y$, then $x_{n_{k_j}} \to y$ as well, a contradiction. So $K$ is not compact.

I think the following is a counterexample.

Consider the Stone–Čech compactification $\beta \mathbb{N}$. Fix some $x \in \beta \mathbb{N} \setminus \mathbb{N}$ and set $S = \beta \mathbb{N} \setminus \{x\}$, which is locally compact Hausdorff and not compact. Let $\mu$ be the Borel measure on $S$ which puts mass $2^{-n}$ on each point $n \in \mathbb{N} \subset S$. This measure is regular, and its support is all of $S$ (which is non-compact), since $\mathbb{N}$ is dense in $S$.

Suppose $\mathcal{H}$ is an infinite disjoint family of Borel sets having positive measure. Then each $H \in \mathcal{H}$ must contain at least one point of $\mathbb{N}$, so write $\mathcal{H} = \{H_1, H_2, \dots\}$, and for each $k$ choose some $n_k \in H_k \cap \mathbb{N}$. Set $A = \{n_1, n_3, n_5, \dots\}$ and $B = \{n_2, n_4, n_6, \dots\}$. I claim that either $A$ or $B$ is contained in a compact set $K$.

To see this, take some function $f : \mathbb{N} \to \{0,1\}$ which is $0$ on $A$ and $1$ on $B$, and extend it to a continuous $\hat{f} : \beta \mathbb{N} \to \{0,1\}$. Suppose that $\hat{f}(x) = 1$. Then the set $K = \hat{f}^{-1}(\{0\})$ is closed in $\beta \mathbb{N}$, hence compact, and contains $A$ but not $x$. So $K$ is also compact in $S$, and since it contains $A$, it intersects $H_1, H_3, H_5, \dots$ with positive measure. If instead we had $\hat{f}(x) = 0$, then just interchange $A$ and $B$, and get a compact set $K$ intersecting $H_2, H_4, H_6,\dots$.

A couple of notes in the positive direction:

• The statement is true if $\mu$ is not $\sigma$-finite. For then, by transfinite induction (or Zorn's lemma), we can find an uncountable collection $\mathbb{H}$ of pairwise disjoint Borel sets with positive measures. Any set $K$ which meets all of them with positive measure must then have measure infinity, and since $\mu$ was regular, such a set cannot be compact.

• The statement is true if $S$ is metrizable. We can suppose without loss of generality that $\mu$ has full support, and by the previous note, we can also suppose that $\mu$ is $\sigma$-finite. Then, by inner regularity of the measure, there is a $\sigma$-compact set $E$ of full measure, which in particular is separable. Since $\mu$ has full support, $E$ is dense, so $S$ is also separable. But a separable metric space is second countable, and a second countable locally compact space is $\sigma$-compact, and you already know the statement holds for $\sigma$-compact spaces.

I think the following is a counterexample.

Consider the Stone–Čech compactification $\beta \mathbb{N}$. Fix some $x \in \beta \mathbb{N} \setminus \mathbb{N}$ and set $S = \beta \mathbb{N} \setminus \{x\}$, which is locally compact Hausdorff and not compact. Let $\mu$ be the Borel measure on $S$ which puts mass $2^{-n}$ on each point $n \in \mathbb{N} \subset S$. This measure is regular, and its support is all of $S$ (which is non-compact), since $\mathbb{N}$ is dense in $S$.

Suppose $\mathcal{H}$ is an infinite disjoint family of Borel sets having positive measure. Then each $H \in \mathcal{H}$ must contain at least one point of $\mathbb{N}$, so write $\mathcal{H} = \{H_1, H_2, \dots\}$, and for each $k$ choose some $n_k \in H_k \cap \mathbb{N}$. Set $A = \{n_1, n_3, n_5, \dots\}$ and $B = \{n_2, n_4, n_6, \dots\}$. I claim that either $A$ or $B$ is contained in a compact set $K$.

To see this, take some function $f : \mathbb{N} \to \{0,1\}$ which is $0$ on $A$ and $1$ on $B$, and extend it to a continuous $\hat{f} : \beta \mathbb{N} \to \{0,1\}$. Suppose that $\hat{f}(x) = 1$. Then the set $K = \hat{f}^{-1}(\{0\})$ is closed in $\beta \mathbb{N}$, hence compact, and contains $A$ but not $x$. So $K$ is also compact in $S$, and since it contains $A$, it intersects $H_1, H_3, H_5, \dots$ with positive measure. If instead we had $\hat{f}(x) = 0$, then just interchange $A$ and $B$, and get a compact set $K$ intersecting $H_2, H_4, H_6,\dots$.