The support of a Borel probability on a separable Banach space has full measure

By [F1], choose an at most countable dense set $C\subseteq X$. Since $\mu (X)=1$ and $\mu (\mathrm{\varnothing })=0$ by [F7], $X$ is nonempty, so $C$ is nonempty. Let ${\mathbb{Q}}_{>0}$ be the positive rationals. By [F3], $C\times {\mathbb{Q}}_{>0}$ is a nonempty at most countable set, so [F4] gives a surjection $e:\mathbb{N}\to C\times {\mathbb{Q}}_{>0}$. The composite of $e$ with $(c,q)\mapsto B(c,q)$ is a surjection onto ${\mathcal{B}}_0:=\{B(c,q):c\in C,q\in {\mathbb{Q}}_{>0}\}$, so ${\mathcal{B}}_0$ is at most countable. Each of its members is open by [F2]. It is a base: if $x\in U$ with $U$ open, choose $r>0$ so $B(x,r)\subseteq U$, choose $c\in C$ with $\mathrm{\parallel }x-c{\mathrm{\parallel }}_X<r\mathrm{/}3$, and choose rational $q$ with $\mathrm{\parallel }x-c{\mathrm{\parallel }}_X<q<r-\mathrm{\parallel }x-c{\mathrm{\parallel }}_X$ using density in [F3]. Then $x\in B(c,q)$, and for every $z\in B(c,q)$ the triangle inequality gives $\mathrm{\parallel }z-x{\mathrm{\parallel }}_X<q+\mathrm{\parallel }x-c{\mathrm{\parallel }}_X<r$, so $x\in B(c,q)\subseteq U$.

Let $\mathcal{N}:=\{B\in {\mathcal{B}}_0:\mu (B)=0\}$ and $N:=\bigcup \mathcal{N}$. By [F3], $\mathcal{N}$ is at most countable; by [F5] $N$ is Borel, and by [F6] its measure is zero (with the empty-family case giving $N=\mathrm{\varnothing }$). If $x\in N$, a zero-measure open basis ball contains $x$, so $x\mathrm{\notin }\mathrm{supp}\mu$. If $x\mathrm{\notin }N$, every open neighborhood $U$ of $x$ contains some $B\in {\mathcal{B}}_0$ with $x\in B\subseteq U$ by step 1.1; since $B\mathrm{\notin }\mathcal{N}$, $\mu (B)>0$, and [F6] gives $\mu (U)\ge \mu (B)>0$. Thus $X\setminus N=\mathrm{supp}\mu$, which is Borel and has measure one.