Why is the following topology of probability measures Hausdorff? Let $X$ be a Hausdorff topological space. Let $PX$ be the set of all Borel probability measures on $X$. Bogachev's Measure Theory (vol. II, section 8.10.iv) defines the $A$-topology on $PX$ to be the one generated by the following opens,
$$
U(\mu, G, \varepsilon) := \{\nu \in PX : \mu(G) < \nu(G) + \varepsilon \},
$$
for all $\mu\in PX$, $G\subseteq X$ open, and $\varepsilon > 0$. 
It then says that since two Borel measures are equal if and only if they coincide on all open sets (Lemma 7.1.2 in there), the $A$-topology is Hausdorff. 
This is not clear to me: how does the proof proceed, in more detail? 
Here is what I figured out so far. Suppose that $\mu,\nu\in PX$ are not equal. In other words, there exists an open set $G\subseteq X$ such that $\mu(G)\ne \nu(G)$. Suppose, for example, that $\mu(G) > \nu(G)$. Then there exists $\varepsilon>0$ such that $\mu(G) \ge \nu(G) + \varepsilon$. Therefore $\nu\notin U(\mu, G, \varepsilon)$, while clearly $\mu\in U(\mu, G, \varepsilon)$. We have found an open neighborhood of $\mu$ that does not contain $\nu$. But how can we find an open neighborhood of $\nu$ that does not contain $\mu$?
 A: This is not true without additional hypotheses.  For instance, let $X=\omega_1+1$, let $\mu$ be the point measure at $\omega_1$ and let $\nu$ be the measure supported on $X\setminus\{\omega_1\}$ corresponding to the club filter.  Then for any open set $G$, $\nu(G)\geq \mu(G)$: this is clear if $\omega_1\not\in G$, and if $\omega_1\in G$ then $\nu(G)$ must be $1$ since $G$ is open.  It follows that every open set which contains $\mu$ also contains $\nu$.
It is true if you require all your measures to be inner regular.  Indeed, in that case, suppose $\mu\neq\nu$ so there is some Borel set $A$ such that $\mu(A)>\nu(A)$.  Then $\mu(A)+\nu(X\setminus A)>1$, so by inner regularity there exist compact $K\subseteq A$ and $L\subseteq X\setminus A$ such that $\mu(K)+\nu(L)>1$.  Since $K$ and $L$ are compact and $X$ is Hausdorff, there exist disjoint open sets $G\supseteq K$ and $H\supseteq L$.  Since $\mu(G)+\nu(H)>1$, for $\epsilon$ sufficiently small, any $m\in U(\mu, G,\epsilon)\cap U(\nu,H,\epsilon)$ will satisfy $m(G)+m(H)>1$, which is impossible.  Thus $U(\mu,G,\epsilon)$ and $U(\nu,H,\epsilon)$ are disjoint open sets separating $\mu$ and $\nu$.
