I know this is true for separable metric spaces, and locally compact metric spaces, but is it true in general?

Following Pietro's lead, let me observe that if there is a measurable cardinal, then there is a counterexample. Suppose that $\kappa$ is a measurable cardinal. Then there is a $\kappa$additive 2valued measure $\mu$, measuring all subsets of $\kappa$, giving them either measure $0$ or $1$, giving measure $1$ to the whole space and giving measure $0$ to any set of size less than $\kappa$ (among others). If we give $\kappa$ the discrete topology, then every set is closed (and hence Borel), and the support is empty. 


Every $\sigma$smooth measure is $\tau$smooth. This is what we need. As noted, if there is a (realvalued) measurable cardinal, then this may fail for a metric space. A space is called "measurecompact" iff every $\sigma$smooth measure is $\tau$smooth. The reference for all of this (up to 1965) is: V. S. Varadarajan, "Measures on Topological Spaces". In a completely regular space we would use "zero sets" (a set where some continuous realvalued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$smooth) this intersection again has measure $1$. As I recall, a metric space is measurecompact if and only if there is no discrete subset with realvalued measurable cardinal. So, in particular, if there are no realvalued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC. The term "measurecompact" is due to Moran, 1965. By analogy with "realcompact" which may be characterized in the same way using only $\{0,1\}$valued measures. 


Consider an uncountable discrete metric space $X $ (i.e., metrized by the Kronecker delta). Define a measure on $X$ putting for any $A\subset X,\\ $ $\mu(A)=1$ or $\mu(A)=0$ according whether $A$ belongs to a given nonprincipal ultrafilter $\mathcal{F}$, or not (sigmaadditivity holds, for there are no disjoint subsets of positive measure). Then $\mu$ is a Borel probability measure with empty support. [edit] Actually, this is additive, but to ensure sigmaadditivity it would be needed that $\mathcal{F}$ be closed under countable intersections. 


Here's a simple argument for why large cardinals are really needed here: Suppose $\kappa$ is the least cardinal such that there is a collection of size $\kappa$ of open null sets with non null union. Let $I$ be the sigmaideal of those subsets of $\kappa$ over which the union of these open sets is null. Then the boolean algebra $\mathcal{P}(\kappa)/I$ satisfies the countable chain condition since otherwise, there would be uncountably many pairwise disjoint non null open sets. Cardinals which admit such ideals are sometimes called quasimeasurable. Using Ulam's matrix, it can be verified that the least quasimeasurable is weakly inaccessible. 


Rather than a counterexample, here is an equivalent condition for topological spaces (and hence metric spaces). Let $(\Lambda,\mathcal{T})$ be a topological space with the Borel $\sigma$algebra and $\mu$ a probability measure on it. Say that $\mathcal{T}$ outlines $\mu$ if given an arbitrary collection of open sets $\mathcal{U}\subset \mathcal{T}$ with $\mu(U)=0$ for every $U \in \mathcal{U}$, $$ \mu\left(\bigcup_{U\in \mathcal{U}}U\right) = 0 $$ Let $S := \mathrm{supp}(\mu)$. Then $\mathcal{T}$ outlines $\mu$ if and only if $\mu(S) = 1$. Proof: ($\Rightarrow$) Observe that $y \in \Lambda \backslash S \iff$ there exists an open neighbourhood $U_y$of $y$ with $\mu(U_y) = 0$, in which case $U_y \subset \Lambda \backslash S$. Taking the union, $\Lambda \backslash S = \bigcup_{y\in \Lambda \backslash S}U_y$, so that $\mu(\Lambda \backslash S)=0$ by assumption. Hence $\mu(S) = 1$ since $\Lambda = S \cup \Lambda \backslash S$ and both sets are certainly Borelmeasurable. ($\Leftarrow$) Suppose that $\mathcal{U}$ is a collection of open subsets of measure zero. Then $S \subset \Lambda \backslash \bigcup \mathcal{U}$, so that bya similar argument to above, $\Lambda = S \cup \Lambda \backslash S \implies \mu(U) \leq 0$, and hence $\mu(U) = 0$ by nonnegativity. So $\mathcal{T}$ outlines $\mu$. This concludes the proof. Note that the condition ``$\mathcal{T}$ outlines $\mu$'' is always satisfied for a measure on a separable metric space, since such spaces are second countable, so the union in $(\Rightarrow)$ can be reduced to a countable union, which gives a measure of zero by the assumed countable additivity of a measure. [I came up with this proof myself, sorry I don't have further sources for you to look at.] 

