Let $X$ be a metric space and $\mu:\mathscr{B}_X\rightarrow [0,\infty]$ be a Borel probability measure.
Define $supp(\mu)$ to be the set of all points whose every open neighborhood has a postive measure.
If $supp(\mu)$ is separable, why is $\mu(supp(\mu))=1$?
(I searched for it and I saw that this is true when $X$ is a metric space, but false in general. I cannot find a proof of it though)
This is false in general. For instance, if $\kappa$ is a measurable cardinal, you can take $X=\kappa$ with the discrete metric and let $\mu$ be the $\{0,1\}$-valued measure corresponding to a nonprincipal $\kappa$-complete ultrafilter. Then $\mu$ vanishes on every singleton so its support is empty.
(Probably you can also cook up a counterexample in just ZFC, but I don't know one off the top of my head.)
It is true if you assume $X$ is separable, because the complement of the support can be covered by open sets of measure $0$ and by separability you can take a countable subcover to conclude that the complement of the support has measure $0$.