Let $(E, d)$ be a metric space and $\mu$ a Borel probability measure on $E$. Assume $\mu$ has a property that if $(E_1, E_2)$ is a measurable partition of $E$, i.e., $E_1 \cap E_2 = \emptyset$ and $E_1 \cup E_2 = E$, then either $\mu(E_1)=0$ or $\mu(E_1)=1$. Let $U$ be the support of $\mu$, i.e., $U$ is the intersection of all closed sets with full measure.
If $E$ is separable, then $\mu(U) = 1$ and thus $U \neq \emptyset$. We fix $a \in U$. Then $(B(a, 1/n), (B(a, 1/n))^c)$ is a measurable partition of $E$, with $B(a, 1/n)$ being the open ball centered at $a$ with radius $1/n$. It follows that $\mu(B(a, 1/n)) > 0$ and thus $\mu(B(a, 1/n)) = 1$ for all $n$. Hence $\mu (\{a\})=1$ and thus $U = \{a\}$.
Do we have the same conclusion, i.e., $\mu$ concentrates on a single point, if we drop the hypothesis that $E$ is separable?