Every $\sigma$-smooth measure is $\tau$-smooth.  This is what we need.  As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space.  A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1960) is: Gillman & Jerison, RINGS OF CONTINUOUS FUNCTIONS.  In a completely regular space we would use zero sets" (a set where some continuous real-valued function vanishes).  Butin 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 measure-compact if and only if there is no discrete subset with real-valued measurable cardinal.  So, in particular, if there are no real-valued 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.