Let $\mu$ be a (positive, probability) measure satisfying the hypothesis. Note that there cannot be any Borel set $A$ with $1/3 \le \mu(A) \le 2/3$, since then in the partition $S = A \cup A^c$, the smaller set would have measure at least $1/3$. Thus it suffices to show there is an $x_0$ with $\mu(\{x_0\}) \ge 1/3$.

Suppose not; then every point has measure less than $1/3$. By regularity, every point thus has an open neighborhood with measure less than $1/3$. By compactness, I can find a finite cover of $S$ by such open sets, $U_1, \dots, U_n$. But if we let $V_k = U_1 \cup \dots \cup U_k$ for $1 \le k \le n$, then $\mu(V_{k+1}) < \mu(V_k) + 1/3$, and $V_n = S$ so that $\mu(V_n) = 1$. By "pigeonhole" some $V_n$ must have measure between $1/3$ and $2/3$ which is a contradiction.

Let $\mu$ be a (positive, probability) measure satisfying the hypothesis. Note that there cannot be any Borel set $A$ with $1/3 \le \mu(A) \le 2/3$, since then in the partition $S = A \cup A^c$, the smaller set would have measure at least $1/3$. Thus it suffices to show there is an $x_0$ with $\mu(\{x_0\}) \ge 1/3$.

Suppose not; then every point has measure less than $1/3$. By regularity, every point thus has an open neighborhood with measure less than $1/3$. By compactness, I can find a finite cover of $S$ by such open sets, $U_1, \dots, U_n$. Let $V_k = U_1 \cup \dots \cup U_k$ for $1 \le k \le n$. Let $m = \max\{ k : \mu(V_k) < 1/3 \}$; in particular $m < n$ (since $V_n = S$). Then $$1/3 \le \mu(V_{m+1}) < \mu(V_m) + 1/3 < 2/3$$ which is a contradiction.