The set $MS=\{(\mu,K)\in P_X\times K_X:\mathrm{supp}(\mu)=K\}$ is of type $G_\delta$ in $P_X\times K_X$ and hence the weak+Hausdorff topology on $P_X$ is Polish.
Indeed, fix any countable base $\{U_n\}_{n\in\mathbb N}$ of the topology of $X$ and observe that $$MS=\bigcap_{n\in\mathbb N}\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}.$$
So, it remains to show that for every $n\in\mathbb N$ the set
$$MS_n:=\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}$$is of type $G_\delta$ in $P_X\times K_X$.
Observe that $(P_X\times K_X)\setminus MS_n=MS_n'\cup MS_n''$ where
$$MS_n':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)>0\;\wedge\;U_n\cap K=\emptyset\}$$ and
$$MS_n'':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)=0\;\wedge\;U_n\cap K\ne\emptyset\}.$$
The known properties of the topologies on the spaces $P_X$ and $K_X$ ensure that the set $\{\mu\in P_X:\mu(U_n)>0\}$ is open in $P_X$ and the set $\{K\in K_X:K\cap U_n\ne \emptyset\}$ is open in $K_X$. Then $MS_n'$ and $M_s''$ are intersections of open and closed sets, so are of type $F_\sigma$ in $P_X\times K_X$. Then $MS_n$ is of type $G_\delta$ in $P_X\times K_X$ and so is the set $MS=\bigcap_{n\in\mathbb N}MS_n$.
$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Take any $\mu\in\M(\Om)$, any open subset $\Th$ of $\Om$, and any real $\ep>0$. Let $\de:=\ep/4$.
By the Hahn decomposition theorem, there is a partition of $\Om$ into Borel sets $D^\pm$ such that $D^+$ is a positive set for $\mu$ and $D^-$ is a negative set for $\mu$.
Let
\begin{equation*}
A^\pm:=\Th\cap D^\pm. \tag{1}\label{1}
\end{equation*}
Since $|\mu|$ is inner regular, there exist compact sets
\begin{equation*}
K^\pm\subseteq A^\pm\text{ such that }|\mu|(A^\pm\setminus K^\pm)<\de. \tag{2}\label{2}
\end{equation*}
Since $\Om$ is normal, there exist open subsets $U^\pm$ of $\Th$ such that
\begin{equation*}
U^\pm\supseteq K^\pm\text{ and }U^+\cap U^-=\emptyset. \tag{3}\label{3}
\end{equation*}
Since the sets $K^\pm$ are compact and $\Om$ is locally compact, without loss of generality the closures of the sets $U^\pm$ are compact.
By Urysohn'slemma, there exist continuous functions $f^\pm\colon\Om\to\R$ such that
\begin{equation*}
0\le f^\pm\le1,\quad f^\pm=1\text{ on }K^\pm,\quad f^\pm=0\text{ on }\Om\setminus U^\pm. \tag{4}\label{4}
\end{equation*}
Let
\begin{equation*}
f:=f^+-f^-.
\end{equation*}
Then $f^+f^-=0$, whence $|f|\le1$. Also, $f=0$ on $\Om\setminus(U^+\cup U^-)$. So, recalling that the closures of the sets $U^\pm$ are compact, we see that $f\in C_c(\Om)$. Also, since $U^\pm$ are subsets of $\Th$, we have $|f|\le1_\Th$.
It remains to show that
\begin{equation*}
\int_\Om f\,d\mu\ge|\mu|(\Th)-\ep. \tag{$*$}\label{*}
\end{equation*}
To do this, note that, by \eqref{3}, \eqref{2}, and \eqref{1},
\begin{equation}
\begin{aligned}
|\mu|(U^-\setminus K^-)&\le|\mu|(\Th\setminus U^+\setminus K^-) \\
&=|\mu|(\Th)-|\mu|(U^+)-|\mu|(K^-) \\
&\le|\mu|(\Th)-|\mu|(K^+)-|\mu|(K^-) \\
&<|\mu|(\Th)-|\mu|(A^+)-|\mu|(A^-)+2\de=2\de.
\end{aligned}
\tag{5}\label{5}
\end{equation}
So, by \eqref{4}, \eqref{3}, \eqref{2}, \eqref{5}, and \eqref{1},
\begin{equation*}
\begin{aligned}
\int_\Om f\,d\mu&=\int_{U^+} f^+\,d\mu-\int_{U^-} f^-\,d\mu \\
&\ge\int_{K^+} f^+\,d\mu-\int_{K^-} f^-\,d\mu -\int_{U^-\setminus K^-} f^-\,d\mu \\
&=\mu(K^+)-\mu(K^-) -\int_{U^-\setminus K^-} f^-\,d\mu \\
&\ge\mu(K^+)-\mu(K^-) -|\mu|(U^-\setminus K^-) \\
&>\mu(A^+)-\de-\mu(A^-)-\de -2\de \\
&=|\mu|(\Th)-4\de=|\mu|(\Th)-\ep,
\end{aligned}
\end{equation*}
so that \eqref{*} is proved. $\quad\Box$
Best Answer
Any $f\in C_c(X)$ can be uniformly approximated by functions $f_n\in C_c^\infty(X)$, say by convolving $f$ with appropriate mollifiers $\psi_n\in C_c^\infty(X)$.
So, your desired conclusion indeed follows.