A tight Gaussian law is carried by a separable closed subspace

For each integer $m\ge 1$, tightness and [A1] give a compact set $K_m\subset B$ with $\mu (K_m)>1-2^{-m}$. Each compact metric space $K_m$ has a countable dense subset: for each integer $r\ge 1$, compactness gives a finite $1\mathrm{/}r$-net, and the union of these finite nets is countable and dense. The union over $m$ of these dense subsets is countable; let $S$ be its closure. Then $S$ is separable, contains every $K_m$, and $\mu (S)\ge \mu (K_m)>1-2^{-m}$ for every $m$, hence $\mu (S)=1$.

Let $E$ be the closed real linear span of $S$. Rational linear combinations of a countable dense subset of $S$ form a countable dense subset of its real linear span, so $E$ is separable; it is closed by construction and therefore Banach, and $\mu (E)=1$. Define $W_E=W$ on the measurable event $\{W\in E\}$ and $W_E=0$ on its null complement. Then $W_E$ is $E$-valued and Borel measurable: each Borel set of $E$ is $E\cap A$ for some ambient Borel $A$, and its preimage is $\{W\in E\}\cap W^{-1}(A)$ together with the null complement if the set contains zero. Its law is the restriction of $\mu$ to $E$. For any $\mathrm{\ell }\in E^{\ast }$, let $p(x)=\mathrm{\parallel }\mathrm{\ell }{\mathrm{\parallel }}_{E^{\ast }}\mathrm{\parallel }x{\mathrm{\parallel }}_B$ on $B$; $\mathrm{\ell }$ is dominated by $p$ on $E$. By [A2] it extends to a linear $\tilde{\mathrm{\ell }}$ on $B$ dominated by $p$, and applying domination to $x$ and $-x$ gives $\mathrm{\mid }\tilde{\mathrm{\ell }}(x)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\ell }{\mathrm{\parallel }}_{E^{\ast }}\mathrm{\parallel }x{\mathrm{\parallel }}_B$. Since $W_E=W$ almost surely, $\mathrm{\ell }(W_E)=\tilde{\mathrm{\ell }}(W)$ almost surely, so $\mathrm{\ell }(W_E)$ is centered normal. Thus the restricted law is Gaussian.

For tightness in $E$, replace each compact $K_m$ by $K_m\cap E$, which is compact and has the same $\mu$-mass because $\mu (E)=1$. If $x\mathrm{\notin }E$, closedness gives a $B$-open neighborhood disjoint from $E$, of $\mu$-mass zero, so $x$ is outside the ambient support. If $x\in E$, every ambient open ball about $x$ intersects $E$ in an $E$-open neighborhood and has the same measure as that intersection; hence all such neighborhoods have positive measure exactly when all relative neighborhoods do. The two supports therefore agree.